{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To run this, press \"*Runtime*\" and press \"*Run all*\" on a **free** Tesla T4 Google Colab instance!\n",
    "<div class=\"align-center\">\n",
    "<a href=\"https://unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/unsloth%20new%20logo.png\" width=\"115\"></a>\n",
    "<a href=\"https://discord.gg/unsloth\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/Discord button.png\" width=\"145\"></a>\n",
    "<a href=\"https://docs.unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/blob/main/images/documentation%20green%20button.png?raw=true\" width=\"125\"></a></a> Join Discord if you need help + \u2b50 <i>Star us on <a href=\"https://github.com/unslothai/unsloth\">Github</a> </i> \u2b50\n",
    "</div>\n",
    "\n",
    "To install Unsloth your local device, follow [our guide](https://docs.unsloth.ai/get-started/install-and-update). This notebook is licensed [LGPL-3.0](https://github.com/unslothai/notebooks?tab=LGPL-3.0-1-ov-file#readme).\n",
    "\n",
    "You will learn how to do [data prep](#Data), how to [train](#Train), how to [run the model](#Inference), & [how to save it](#Save)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### News"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "Introducing FP8 precision training for faster RL inference. [Read Blog](https://docs.unsloth.ai/new/fp8-reinforcement-learning).\n",
    "\n",
    "Unsloth's [Docker image](https://hub.docker.com/r/unsloth/unsloth) is here! Start training with no setup & environment issues. [Read our Guide](https://docs.unsloth.ai/new/how-to-train-llms-with-unsloth-and-docker).\n",
    "\n",
    "[gpt-oss RL](https://docs.unsloth.ai/new/gpt-oss-reinforcement-learning) is now supported with the fastest inference & lowest VRAM. Try our [new notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/gpt-oss-(20B)-GRPO.ipynb) which creates kernels!\n",
    "\n",
    "Introducing [Vision](https://docs.unsloth.ai/new/vision-reinforcement-learning-vlm-rl) and [Standby](https://docs.unsloth.ai/basics/memory-efficient-rl) for RL! Train Qwen, Gemma etc. VLMs with GSPO - even faster with less VRAM.\n",
    "\n",
    "Visit our docs for all our [model uploads](https://docs.unsloth.ai/get-started/all-our-models) and [notebooks](https://docs.unsloth.ai/get-started/unsloth-notebooks).\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Installation"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": "%%capture\nimport os\nos.environ[\"UNSLOTH_VLLM_STANDBY\"] = \"1\" # [NEW] Extra 30% context lengths!\n!pip install --upgrade -qqq uv\ntry: import numpy, PIL; get_numpy = f\"numpy=={numpy.__version__}\"; get_pil = f\"pillow=={PIL.__version__}\"\nexcept: get_numpy = \"numpy\"; get_pil = \"pillow\"\ntry: import subprocess; is_t4 = \"Tesla T4\" in str(subprocess.check_output([\"nvidia-smi\"]))\nexcept: is_t4 = False\nget_vllm, get_triton = (\"vllm==0.9.2\", \"triton==3.2.0\") if is_t4 else (\"vllm==0.10.2\", \"triton\")\n!uv pip install -qqq --upgrade     unsloth {get_vllm} {get_numpy} {get_pil} torchvision bitsandbytes xformers\n!uv pip install -qqq {get_triton}\n!uv pip install \"huggingface_hub>=0.34.0\" \"datasets==4.3.0\n!uv pip install transformers==4.56.2\n!uv pip install --no-deps trl==0.22.2"
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "ZkH_y8UC9lvv"
   },
   "source": [
    "### Unsloth"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "jN75nmdx9lvw"
   },
   "source": [
    "Goal: To convert `Llama-3.2-1B-Instruct` into a reasoning model via GRPO by using OpenR1's Math dataset.\n",
    "\n",
    "We first pre fine-tune the model to make GRPO skip trying to match formatting - this speeds GRPO up."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 1000,
     "referenced_widgets": [
      "3f0c6bb2c4e34868ae8279ee2a8a50eb",
      "c01279407be8434a9d441f98927ec9c9",
      "e1f06e01d9f04cca9cd2909e80af936b",
      "920dd87c82dd48ca8f82091d74039850",
      "139a54bc7ed9446eb33b5df37ca7588e",
      "bdc878527ae04540ac1d82df32e13fb8",
      "a5c83b8ef546449fb951769f49748419",
      "0dfc56ea5cd54660bba22b7cd632d2f9",
      "9d01e1ad71ad437b930d7f7c059a91d2",
      "8b64abbe1a2648a090c472a869f44a4f",
      "ac87ede4cdf646e2b166badd2d8f0dea"
     ]
    },
    "id": "DkIvEkIIkEyB",
    "outputId": "5b44e72d-7957-46a6-b711-93648b06bc4d"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\ud83e\udda5 Unsloth: Will patch your computer to enable 2x faster free finetuning.\n",
      "\ud83e\udda5 Unsloth Zoo will now patch everything to make training faster!\n",
      "INFO 11-25 15:07:49 [vllm_utils.py:702] Unsloth: Patching vLLM v1 graph capture\n",
      "==((====))==  Unsloth 2025.11.3: Fast Llama patching. Transformers: 4.57.1. vLLM: 0.11.2.\n",
      "   \\\\   /|    NVIDIA L4. Num GPUs = 1. Max memory: 22.161 GB. Platform: Linux.\n",
      "O^O/ \\_/ \\    Torch: 2.9.0+cu126. CUDA: 8.9. CUDA Toolkit: 12.6. Triton: 3.5.0\n",
      "\\        /    Bfloat16 = TRUE. FA [Xformers = 0.0.33.post1. FA2 = False]\n",
      " \"-____-\"     Free license: http://github.com/unslothai/unsloth\n",
      "Unsloth: Fast downloading is enabled - ignore downloading bars which are red colored!\n",
      "Unsloth: vLLM loading unsloth/Llama-3.2-1B-Instruct-FP8-Block with actual GPU utilization = 86.62%\n",
      "Unsloth: Your GPU has CUDA compute capability 8.9 with VRAM = 22.16 GB.\n",
      "Unsloth: Using conservativeness = 1.0. Chunked prefill tokens = 2048. Num Sequences = 288.\n",
      "Unsloth: vLLM's KV Cache can use up to 17.53 GB. Also swap space = 4 GB.\n",
      "Unsloth: Not an error, but `use_cudagraph` is not supported in vLLM.config.CompilationConfig. Skipping.\n",
      "WARNING 11-25 15:07:58 [compilation.py:610] Level is deprecated and will be removed in the next release,either 0.12.0 or 0.11.2 whichever is soonest.Use mode instead.If both level and mode are given,only mode will be used.\n",
      "WARNING 11-25 15:07:58 [compilation.py:699] The 'use_inductor' flag is deprecated and will be removed in the next release (v0.12.0). Please use the 'backend' option instead.\n",
      "Unsloth: Not an error, but `device` is not supported in vLLM. Skipping.\n",
      "{'model': 'unsloth/Llama-3.2-1B-Instruct-FP8-Block', 'gpu_memory_utilization': 0.8661508568217998, 'max_model_len': 2048, 'quantization': None, 'load_format': 'auto', 'kv_cache_dtype': 'auto', 'dtype': torch.bfloat16, 'max_num_batched_tokens': 2048, 'max_num_seqs': 288, 'max_logprobs': 0, 'seed': 0, 'enable_lora': True, 'max_lora_rank': 32, 'max_loras': 1, 'disable_log_stats': True, 'enable_prefix_caching': True, 'enable_chunked_prefill': True, 'compilation_config': {'level': 3, 'mode': 3, 'debug_dump_path': None, 'cache_dir': '', 'compile_cache_save_format': 'binary', 'backend': 'inductor', 'custom_ops': [], 'splitting_ops': None, 'compile_mm_encoder': False, 'use_inductor': True, 'compile_sizes': None, 'inductor_compile_config': {'epilogue_fusion': True, 'max_autotune': False, 'shape_padding': True, 'trace.enabled': False, 'triton.cudagraphs': False, 'debug': False, 'dce': True, 'memory_planning': True, 'coordinate_descent_tuning': False, 'trace.graph_diagram': False, 'compile_threads': 12, 'group_fusion': True, 'disable_progress': False, 'verbose_progress': True, 'triton.multi_kernel': 0, 'triton.use_block_ptr': True, 'triton.enable_persistent_tma_matmul': True, 'triton.autotune_at_compile_time': False, 'triton.cooperative_reductions': False, 'cuda.compile_opt_level': '-O2', 'cuda.enable_cuda_lto': True, 'combo_kernels': False, 'benchmark_combo_kernel': True, 'combo_kernel_foreach_dynamic_shapes': True, 'enable_auto_functionalized_v2': False}, 'inductor_passes': {}, 'cudagraph_mode': <CUDAGraphMode.FULL_AND_PIECEWISE: (2, 1)>, 'cudagraph_num_of_warmups': 1, 'cudagraph_capture_sizes': None, 'cudagraph_copy_inputs': False, 'cudagraph_specialize_lora': True, 'use_inductor_graph_partition': False, 'pass_config': {}, 'max_cudagraph_capture_size': None, 'local_cache_dir': None}, 'enforce_eager': False, 'swap_space': 4, 'enable_sleep_mode': True, 'disable_cascade_attn': False}\n",
      "INFO 11-25 15:07:58 [utils.py:253] non-default args: {'dtype': torch.bfloat16, 'seed': 0, 'max_model_len': 2048, 'enable_prefix_caching': True, 'gpu_memory_utilization': 0.8661508568217998, 'max_num_batched_tokens': 2048, 'max_num_seqs': 288, 'max_logprobs': 0, 'disable_log_stats': True, 'enable_lora': True, 'max_lora_rank': 32, 'enable_chunked_prefill': True, 'compilation_config': {'level': 3, 'mode': 3, 'debug_dump_path': None, 'cache_dir': '', 'compile_cache_save_format': 'binary', 'backend': 'inductor', 'custom_ops': [], 'splitting_ops': None, 'compile_mm_encoder': False, 'use_inductor': True, 'compile_sizes': None, 'inductor_compile_config': {'epilogue_fusion': True, 'max_autotune': False, 'shape_padding': True, 'trace.enabled': False, 'triton.cudagraphs': False, 'debug': False, 'dce': True, 'memory_planning': True, 'coordinate_descent_tuning': False, 'trace.graph_diagram': False, 'compile_threads': 12, 'group_fusion': True, 'disable_progress': False, 'verbose_progress': True, 'triton.multi_kernel': 0, 'triton.use_block_ptr': True, 'triton.enable_persistent_tma_matmul': True, 'triton.autotune_at_compile_time': False, 'triton.cooperative_reductions': False, 'cuda.compile_opt_level': '-O2', 'cuda.enable_cuda_lto': True, 'combo_kernels': False, 'benchmark_combo_kernel': True, 'combo_kernel_foreach_dynamic_shapes': True, 'enable_auto_functionalized_v2': False}, 'inductor_passes': {}, 'cudagraph_mode': <CUDAGraphMode.FULL_AND_PIECEWISE: (2, 1)>, 'cudagraph_num_of_warmups': 1, 'cudagraph_capture_sizes': None, 'cudagraph_copy_inputs': False, 'cudagraph_specialize_lora': True, 'use_inductor_graph_partition': False, 'pass_config': {}, 'max_cudagraph_capture_size': None, 'local_cache_dir': None}, 'enable_sleep_mode': True, 'model': 'unsloth/Llama-3.2-1B-Instruct-FP8-Block'}\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "/usr/local/lib/python3.12/dist-packages/pydantic/type_adapter.py:605: UserWarning: Pydantic serializer warnings:\n",
      "  PydanticSerializationUnexpectedValue(Expected `enum` - serialized value may not be as expected [field_name='mode', input_value=3, input_type=int])\n",
      "  return self.serializer.to_python(\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:07:59 [model.py:631] Resolved architecture: LlamaForCausalLM\n",
      "INFO 11-25 15:07:59 [model.py:1745] Using max model len 2048\n",
      "INFO 11-25 15:08:01 [scheduler.py:216] Chunked prefill is enabled with max_num_batched_tokens=2048.\n",
      "INFO 11-25 15:08:02 [core.py:93] Initializing a V1 LLM engine (v0.11.2) with config: model='unsloth/Llama-3.2-1B-Instruct-FP8-Block', speculative_config=None, tokenizer='unsloth/Llama-3.2-1B-Instruct-FP8-Block', skip_tokenizer_init=False, tokenizer_mode=auto, revision=None, tokenizer_revision=None, trust_remote_code=False, dtype=torch.bfloat16, max_seq_len=2048, download_dir=None, load_format=auto, tensor_parallel_size=1, pipeline_parallel_size=1, data_parallel_size=1, disable_custom_all_reduce=False, quantization=compressed-tensors, enforce_eager=False, kv_cache_dtype=auto, device_config=cuda, structured_outputs_config=StructuredOutputsConfig(backend='auto', disable_fallback=False, disable_any_whitespace=False, disable_additional_properties=False, reasoning_parser='', reasoning_parser_plugin='', enable_in_reasoning=False), observability_config=ObservabilityConfig(show_hidden_metrics_for_version=None, otlp_traces_endpoint=None, collect_detailed_traces=None), seed=0, served_model_name=unsloth/Llama-3.2-1B-Instruct-FP8-Block, enable_prefix_caching=True, enable_chunked_prefill=True, pooler_config=None, compilation_config={'level': 3, 'mode': 3, 'debug_dump_path': None, 'cache_dir': '', 'compile_cache_save_format': 'binary', 'backend': 'inductor', 'custom_ops': ['none', '+quant_fp8'], 'splitting_ops': ['vllm::unified_attention', 'vllm::unified_attention_with_output', 'vllm::unified_mla_attention', 'vllm::unified_mla_attention_with_output', 'vllm::mamba_mixer2', 'vllm::mamba_mixer', 'vllm::short_conv', 'vllm::linear_attention', 'vllm::plamo2_mamba_mixer', 'vllm::gdn_attention_core', 'vllm::kda_attention', 'vllm::sparse_attn_indexer'], 'compile_mm_encoder': False, 'use_inductor': True, 'compile_sizes': [], 'inductor_compile_config': {'epilogue_fusion': True, 'max_autotune': False, 'shape_padding': True, 'trace.enabled': False, 'triton.cudagraphs': False, 'debug': False, 'dce': True, 'memory_planning': True, 'coordinate_descent_tuning': False, 'trace.graph_diagram': False, 'compile_threads': 12, 'group_fusion': True, 'disable_progress': False, 'verbose_progress': True, 'triton.multi_kernel': 0, 'triton.use_block_ptr': True, 'triton.enable_persistent_tma_matmul': True, 'triton.autotune_at_compile_time': False, 'triton.cooperative_reductions': False, 'cuda.compile_opt_level': '-O2', 'cuda.enable_cuda_lto': True, 'combo_kernels': False, 'benchmark_combo_kernel': True, 'combo_kernel_foreach_dynamic_shapes': True, 'enable_auto_functionalized_v2': False}, 'inductor_passes': {}, 'cudagraph_mode': <CUDAGraphMode.FULL_AND_PIECEWISE: (2, 1)>, 'cudagraph_num_of_warmups': 1, 'cudagraph_capture_sizes': [1, 2, 4, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 272, 288, 304, 320, 336, 352, 368, 384, 400, 416, 432, 448, 464, 480, 496, 512], 'cudagraph_copy_inputs': False, 'cudagraph_specialize_lora': True, 'use_inductor_graph_partition': False, 'pass_config': {}, 'max_cudagraph_capture_size': 512, 'local_cache_dir': None}\n",
      "INFO 11-25 15:08:03 [parallel_state.py:1208] world_size=1 rank=0 local_rank=0 distributed_init_method=tcp://172.28.0.12:56023 backend=nccl\n",
      "INFO 11-25 15:08:03 [parallel_state.py:1394] rank 0 in world size 1 is assigned as DP rank 0, PP rank 0, TP rank 0, EP rank 0\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "/usr/local/lib/python3.12/dist-packages/pydantic/type_adapter.py:605: UserWarning: Pydantic serializer warnings:\n",
      "  PydanticSerializationUnexpectedValue(Expected `enum` - serialized value may not be as expected [field_name='mode', input_value=3, input_type=int])\n",
      "  return self.serializer.to_python(\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:08:03 [topk_topp_sampler.py:36] Using FlashInfer for top-p & top-k sampling.\n",
      "INFO 11-25 15:08:03 [gpu_model_runner.py:3259] Starting to load model unsloth/Llama-3.2-1B-Instruct-FP8-Block...\n",
      "INFO 11-25 15:08:04 [cuda.py:377] Using AttentionBackendEnum.FLASHINFER backend.\n",
      "INFO 11-25 15:08:05 [weight_utils.py:481] No model.safetensors.index.json found in remote.\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "3f0c6bb2c4e34868ae8279ee2a8a50eb",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Loading safetensors checkpoint shards:   0% Completed | 0/1 [00:00<?, ?it/s]\n"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:08:05 [default_loader.py:314] Loading weights took 0.48 seconds\n",
      "INFO 11-25 15:08:06 [punica_selector.py:20] Using PunicaWrapperGPU.\n",
      "INFO 11-25 15:08:07 [gpu_model_runner.py:3338] Model loading took 1.4721 GiB memory and 1.841706 seconds\n",
      "INFO 11-25 15:08:17 [backends.py:631] Using cache directory: /root/.cache/vllm/torch_compile_cache/fbb9cbcbfe/rank_0_0/backbone for vLLM's torch.compile\n",
      "INFO 11-25 15:08:17 [backends.py:647] Dynamo bytecode transform time: 9.86 s\n",
      "INFO 11-25 15:08:20 [backends.py:210] Directly load the compiled graph(s) for dynamic shape from the cache, took 2.227 s\n",
      "WARNING 11-25 15:08:22 [fp8_utils.py:777] Using default W8A8 Block FP8 kernel config. Performance might be sub-optimal! Config file not found at /usr/local/lib/python3.12/dist-packages/vllm/model_executor/layers/quantization/utils/configs/N=3072,K=2048,device_name=NVIDIA_L4,dtype=fp8_w8a8,block_shape=[128,128].json\n",
      "WARNING 11-25 15:08:22 [fp8_utils.py:777] Using default W8A8 Block FP8 kernel config. Performance might be sub-optimal! Config file not found at /usr/local/lib/python3.12/dist-packages/vllm/model_executor/layers/quantization/utils/configs/N=2048,K=2048,device_name=NVIDIA_L4,dtype=fp8_w8a8,block_shape=[128,128].json\n",
      "WARNING 11-25 15:08:22 [fp8_utils.py:777] Using default W8A8 Block FP8 kernel config. Performance might be sub-optimal! Config file not found at /usr/local/lib/python3.12/dist-packages/vllm/model_executor/layers/quantization/utils/configs/N=16384,K=2048,device_name=NVIDIA_L4,dtype=fp8_w8a8,block_shape=[128,128].json\n",
      "WARNING 11-25 15:08:22 [fp8_utils.py:777] Using default W8A8 Block FP8 kernel config. Performance might be sub-optimal! Config file not found at /usr/local/lib/python3.12/dist-packages/vllm/model_executor/layers/quantization/utils/configs/N=2048,K=8192,device_name=NVIDIA_L4,dtype=fp8_w8a8,block_shape=[128,128].json\n",
      "INFO 11-25 15:08:22 [monitor.py:34] torch.compile takes 12.09 s in total\n",
      "INFO 11-25 15:08:23 [gpu_worker.py:359] Available KV cache memory: 17.20 GiB\n",
      "INFO 11-25 15:08:24 [kv_cache_utils.py:1229] GPU KV cache size: 563,744 tokens\n",
      "INFO 11-25 15:08:24 [kv_cache_utils.py:1234] Maximum concurrency for 2,048 tokens per request: 275.27x\n",
      "INFO 11-25 15:08:24 [kernel_warmup.py:65] Warming up FlashInfer attention.\n",
      "INFO 11-25 15:08:24 [vllm_utils.py:707] Unsloth: Running patched vLLM v1 `capture_model`.\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "\rCapturing CUDA graphs (mixed prefill-decode, PIECEWISE):   0%|          | 0/102 [00:00<?, ?it/s]"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "WARNING 11-25 15:08:24 [utils.py:250] Using default LoRA kernel configs\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "Capturing CUDA graphs (mixed prefill-decode, PIECEWISE): 100%|\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588| 102/102 [00:07<00:00, 13.08it/s]\n",
      "Capturing CUDA graphs (decode, FULL): 100%|\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588\u2588| 74/74 [00:05<00:00, 12.83it/s]"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:08:38 [gpu_model_runner.py:4244] Graph capturing finished in 14 secs, took 1.15 GiB\n",
      "INFO 11-25 15:08:38 [vllm_utils.py:714] Unsloth: Patched vLLM v1 graph capture finished in 14 secs.\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:08:39 [core.py:250] init engine (profile, create kv cache, warmup model) took 32.46 seconds\n",
      "INFO 11-25 15:08:41 [llm.py:352] Supported tasks: ('generate',)\n",
      "Unsloth: Just some info: will skip parsing ['k_norm', 'post_feedforward_layernorm', 'layer_norm2', 'post_layernorm', 'norm2', 'norm', 'q_norm', 'layer_norm1', 'input_layernorm', 'attention_norm', 'norm1', 'pre_feedforward_layernorm', 'ffn_norm', 'post_attention_layernorm']\n",
      "Performing substitution for additional_keys=set()\n",
      "Unsloth: Just some info: will skip parsing ['cross_attn_post_attention_layernorm', 'k_norm', 'post_feedforward_layernorm', 'cross_attn_input_layernorm', 'layer_norm2', 'post_layernorm', 'norm2', 'norm', 'q_norm', 'layer_norm1', 'input_layernorm', 'attention_norm', 'norm1', 'pre_feedforward_layernorm', 'ffn_norm', 'post_attention_layernorm']\n"
     ]
    },
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "Unsloth 2025.11.3 patched 16 layers with 16 QKV layers, 16 O layers and 16 MLP layers.\n"
     ]
    }
   ],
   "source": [
    "import os\n",
    "os.environ['UNSLOTH_VLLM_STANDBY'] = \"1\" # Unsloth Standby reduces VRAM by 30%+\n",
    "from unsloth import FastLanguageModel\n",
    "import torch\n",
    "max_seq_length = 2048 # Can increase for longer reasoning traces\n",
    "lora_rank = 32 # Larger rank = smarter, but slower\n",
    "\n",
    "model, tokenizer = FastLanguageModel.from_pretrained(\n",
    "    model_name = \"unsloth/Llama-3.2-1B-Instruct\",\n",
    "    max_seq_length = max_seq_length,\n",
    "    load_in_4bit = False, # False for LoRA 16bit\n",
    "    fast_inference = True, # Enable vLLM fast inference\n",
    "    max_lora_rank = lora_rank,\n",
    "    load_in_fp8 = True, # Float8 RL / GRPO!\n",
    ")\n",
    "\n",
    "model = FastLanguageModel.get_peft_model(\n",
    "    model,\n",
    "    r = lora_rank, # Choose any number > 0 ! Suggested 8, 16, 32, 64, 128\n",
    "    target_modules = [\n",
    "        \"q_proj\", \"k_proj\", \"v_proj\", \"o_proj\",\n",
    "        \"gate_proj\", \"up_proj\", \"down_proj\",\n",
    "    ],\n",
    "    lora_alpha = lora_rank*2, # *2 speeds up training\n",
    "    use_gradient_checkpointing = \"unsloth\", # Reduces memory usage\n",
    "    random_state = 3407,\n",
    ")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "6mojUOPYE_1x",
    "outputId": "00c29dcb-88ac-4ba3-f2d7-bf73284149f6"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<|begin_of_text|><|start_header_id|>system<|end_header_id|>\n",
      "\n",
      "Cutting Knowledge Date: December 2023\n",
      "Today Date: 25 Nov 2025\n",
      "\n",
      "<|eot_id|><|start_header_id|>user<|end_header_id|>\n",
      "\n",
      "Solve x^5 + 3x^4 - 10 = 3.<|eot_id|><|start_header_id|>assistant<|end_header_id|>\n",
      "\n",
      "To solve the equation x^5 + 3x^4 - 10 = 3, we need to isolate the variable x.\n",
      "\n",
      "First, let's add 10 to both sides of the equation:\n",
      "\n",
      "x^5 + 3x^4 = 3 + 10\n",
      "x^5 + 3x^4 = 13\n",
      "\n",
      "Next, let's subtract 13 from both sides of the equation:\n",
      "\n",
      "x^5 + 3x^4 - 13 = 0\n",
      "\n",
      "Now, we have a fifth-degree polynomial equation. Unfortunately, it's not possible to solve this equation analytically (i.e., we can't find an exact solution using standard algebraic techniques). However, we can try to find an approximate solution using numerical methods.\n",
      "\n",
      "One way to do this is to use the Newton-Raphson method, which is a powerful tool for finding roots of real-valued functions. The Newton-Raphson method is based on the idea of iteratively improving an initial guess for the root of the function.\n",
      "\n",
      "Here's an example of how to use the Newton-Raphson method to find an approximate solution:\n",
      "\n",
      "1. Choose an initial guess for the root, x0. For example, let's choose x0 = 1.\n",
      "2. Calculate the derivative of the function at x0, f'(x0). For our equation, f'(x) = 5x^4 + 12x^3.\n",
      "3. Calculate the value of the function at x0, f(x0). For our equation, f(x) = x^5 + 3x^4 - 10.\n",
      "4. Use the Newton-Raphson formula to update the estimate of the root: x1 = x0 - f(x0)/f'(x0).\n",
      "5. Repeat steps 2-4 until the estimate of the root converges to a stable value.\n",
      "\n",
      "Using the Newton-Raphson method, we get:\n",
      "\n",
      "x1 = 1 - (1 - 13)/5\n",
      "x1 = 1 - 12/5\n",
      "x1 = 1 - 2.4\n",
      "x1 = 0.6\n",
      "\n",
      "x2 = 0.6 - (0.6 - 13)/12\n",
      "x2 = 0.6 - (-12.4)/12\n",
      "x2 = 0.6 + 1.04\n",
      "x2 = 1.64\n",
      "\n",
      "x3 = 1.64 - (1.64 - 13)/12\n",
      "x3 =\n"
     ]
    }
   ],
   "source": [
    "messages = [\n",
    "    {\"role\": \"user\", \"content\": \"Solve x^5 + 3x^4 - 10 = 3.\"},\n",
    "]\n",
    "inputs = tokenizer.apply_chat_template(\n",
    "    messages,\n",
    "    add_generation_prompt = True,\n",
    "    return_tensors = \"pt\",\n",
    "    return_dict = True,\n",
    "    reasoning_effort = \"low\",\n",
    ").to(model.device)\n",
    "from transformers import TextStreamer\n",
    "_ = model.generate(**inputs, max_new_tokens = 512, use_cache = True, streamer = TextStreamer(tokenizer))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "oJI8S_c70B3R"
   },
   "source": [
    "Let's call the model as is:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 186,
     "referenced_widgets": [
      "fd98f33a7c584963af768ea61a0c0752",
      "60a3dafa06e84e5f9e96ba65d95e2928",
      "ad1928b451e741bdb885314fe353b0a5",
      "2542a7298c63413aa02d058802fcf6e2",
      "ae86e4d2bc49456395d0ad8562476b79",
      "1bceb4176f8c42ad9e4d6b07aa1377a9",
      "184963b630404e4bb3bbd179da20d6d4",
      "995dec9c6ae7487da1e29b1d92ee9141",
      "e89d34c053344f2580b435b3346de4b3",
      "be65047b5c1c4134aefb8e8f40774629",
      "2219d3900c91470d93eb00557a5211c7",
      "e79fdae492ad437c844a3214ea11b9ba",
      "a6f0d8cb4c804c1bba2c920adc5a56be",
      "2346bbe5f6f84f4f9a1eec75ec463d42",
      "dc7e3c94f84d402fb58ab79095a35307",
      "e2bbd0ede5bd429bb85841d8b2f57bed",
      "2c7af5e97fee44b6871c753cea6853b3",
      "72176176770742e7bee5d6d76697b710",
      "8fcf7030d3344dc0af1c87924779045a",
      "f6bd79fc98b346b0829ceb8ed297384f",
      "9e49b6afb03046ff9afd2a5d5a91ff64",
      "230efdc2fcbd4740bd4fc3b93d4790eb"
     ]
    },
    "id": "aXe9jd0hFBj8",
    "outputId": "8a42f34f-9053-4ba0-b477-b2124d070f1a"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "WARNING 11-25 15:09:22 [abstract.py:312] Executor is not sleeping.\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "fd98f33a7c584963af768ea61a0c0752",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Adding requests:   0%|          | 0/1 [00:00<?, ?it/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "e79fdae492ad437c844a3214ea11b9ba",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "' in the form a^2\\nA) 10.1\\nB) 10.1\\nC) 10.1\\nD) 10.1\\nE) 10.1\\nF) 10.1\\nG) 10.1\\nH) 10.1\\nI) 10.1\\nJ) 10.1\\nI) 10.1\\nJ) 10.1\\nH) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI) 10.1\\nI'"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "text = \"What is the sqrt of 101?\"\n",
    "\n",
    "from vllm import SamplingParams\n",
    "sampling_params = SamplingParams(\n",
    "    temperature = 1.0,\n",
    "    top_k = 50,\n",
    "    max_tokens = 512,\n",
    ")\n",
    "output = model.fast_generate(\n",
    "    [text],\n",
    "    sampling_params = sampling_params,\n",
    "    lora_request = None,\n",
    ")[0].outputs[0].text\n",
    "\n",
    "output"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "W9DuiVRLhMco"
   },
   "source": [
    "### GRPO chat template\n",
    "Since we're using a base model, we should set a chat template. You can make your own chat template as well!\n",
    "1. DeepSeek uses `<think>` and `</think>`, but this is **not** necessary - you can customize it however you like!\n",
    "2. A `system_prompt` is recommended to at least guide the model's responses."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 52
    },
    "id": "6UjowCbT-cFz",
    "outputId": "3640d56d-573c-415d-e2c3-f10726177e1b"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "'You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION>'"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "reasoning_start = \"<start_working_out>\" # Acts as <think>\n",
    "reasoning_end   = \"<end_working_out>\"   # Acts as </think>\n",
    "solution_start  = \"<SOLUTION>\"\n",
    "solution_end    = \"</SOLUTION>\"\n",
    "\n",
    "system_prompt = \\\n",
    "f\"\"\"You are given a problem.\n",
    "Think about the problem and provide your working out.\n",
    "Place it between {reasoning_start} and {reasoning_end}.\n",
    "Then, provide your solution between {solution_start}{solution_end}\"\"\"\n",
    "system_prompt"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "zGgs0MJkDkYL"
   },
   "source": [
    "We create a simple chat template below. Notice `add_generation_prompt` includes prepending `<start_working_out>` to guide the model to start its reasoning process."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "id": "Y3fF9gMujY02"
   },
   "outputs": [],
   "source": [
    "chat_template = \\\n",
    "    \"{% if messages[0]['role'] == 'system' %}\"\\\n",
    "        \"{{ messages[0]['content'] + eos_token }}\"\\\n",
    "        \"{% set loop_messages = messages[1:] %}\"\\\n",
    "    \"{% else %}\"\\\n",
    "        \"{{ '{system_prompt}' + eos_token }}\"\\\n",
    "        \"{% set loop_messages = messages %}\"\\\n",
    "    \"{% endif %}\"\\\n",
    "    \"{% for message in loop_messages %}\"\\\n",
    "        \"{% if message['role'] == 'user' %}\"\\\n",
    "            \"{{ message['content'] }}\"\\\n",
    "        \"{% elif message['role'] == 'assistant' %}\"\\\n",
    "            \"{{ message['content'] + eos_token }}\"\\\n",
    "        \"{% endif %}\"\\\n",
    "    \"{% endfor %}\"\\\n",
    "    \"{% if add_generation_prompt %}{{ '{reasoning_start}' }}\"\\\n",
    "    \"{% endif %}\"\n",
    "\n",
    "# Replace with out specific template:\n",
    "chat_template = chat_template\\\n",
    "    .replace(\"'{system_prompt}'\",   f\"'{system_prompt}'\")\\\n",
    "    .replace(\"'{reasoning_start}'\", f\"'{reasoning_start}'\")\n",
    "tokenizer.chat_template = chat_template"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "vEcLdymBEHdk"
   },
   "source": [
    "Let's see how our chat template behaves on an example:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 70
    },
    "id": "BciEDYSSYFNj",
    "outputId": "32775031-503e-486a-bdb5-44d75c67e8ba"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "\"You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION><|eot_id|>What is 1+1?<start_working_out>I think it's 2.<end_working_out><SOLUTION>2</SOLUTION><|eot_id|>What is 2+2?<start_working_out>\""
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "tokenizer.apply_chat_template([\n",
    "    {\"role\" : \"user\", \"content\" : \"What is 1+1?\"},\n",
    "    {\"role\" : \"assistant\", \"content\" : f\"{reasoning_start}I think it's 2.{reasoning_end}{solution_start}2{solution_end}\"},\n",
    "    {\"role\" : \"user\", \"content\" : \"What is 2+2?\"},\n",
    "], tokenize = False, add_generation_prompt = True)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "_mdsuGjxHrjT"
   },
   "source": [
    "### Pre fine-tuning for formatting\n",
    "We now use a subset of NVIDIA's [Open Math Reasoning dataset](https://huggingface.co/datasets/nvidia/OpenMathReasoning) which was filtered to only include high quality DeepSeek R1 traces.\n",
    "\n",
    "We'll only filter ~59 or so examples to first \"prime\" / pre fine-tune the model to understand our custom GRPO formatting."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 424
    },
    "id": "AXxM2lStVIkd",
    "outputId": "72c0c91f-f77d-4612-b89e-ba8f4e5bf199"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "summary": "{\n  \"name\": \"dataset\",\n  \"rows\": 7507,\n  \"fields\": [\n    {\n      \"column\": \"expected_answer\",\n      \"properties\": {\n        \"dtype\": \"category\",\n        \"num_unique_values\": 868,\n        \"samples\": [\n          \"672\",\n          \"335\",\n          \"575757\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    },\n    {\n      \"column\": \"problem\",\n      \"properties\": {\n        \"dtype\": \"string\",\n        \"num_unique_values\": 3895,\n        \"samples\": [\n          \"A club with 7 members forms three-person committees, but no two committees can have more than one member in common. What is the maximum number of committees that can be formed?\",\n          \"Find the smallest integer \\\\( a > 2 \\\\) such that \\\\( 2 \\\\mid a \\\\), \\\\( 3 \\\\mid (a+1) \\\\), \\\\( 4 \\\\mid (a+2) \\\\), \\\\( 5 \\\\mid (a+3) \\\\), and \\\\( 6 \\\\mid (a+4) \\\\).\",\n          \"Given the polynomial equation \\\\(x^3 - x = -1\\\\) with roots \\\\(a\\\\), \\\\(b\\\\), and \\\\(c\\\\), find the value of \\\\(\\\\frac{1}{1+a} + \\\\frac{1}{1+b} + \\\\frac{1}{1+c}\\\\).\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    },\n    {\n      \"column\": \"generated_solution\",\n      \"properties\": {\n        \"dtype\": \"string\",\n        \"num_unique_values\": 7507,\n        \"samples\": [\n          \"<think>\\nOkay, let's see. I need to solve this problem where there are three prime numbers p, q, and r. The equations given are pq + qr + rp = 191 and p + q = r - 1. The goal is to find p + q + r. Hmm, primes, so they must be 2 or odd primes. Let me start by analyzing the problem step by step.\\n\\nFirst, the equation p + q = r - 1. If I can express r in terms of p and q, maybe I can substitute that into the first equation. Let's try that. So, from the second equation, r = p + q + 1. That seems straightforward. Now, substituting this into the first equation, we get pq + qr + rp = 191. Let's replace r with (p + q + 1).\\n\\nSo, substituting, the first equation becomes:\\n\\npq + q(p + q + 1) + p(p + q + 1) = 191.\\n\\nLet me expand each term:\\n\\nFirst term: pq.\\n\\nSecond term: q*(p + q + 1) = pq + q\\u00b2 + q.\\n\\nThird term: p*(p + q + 1) = p\\u00b2 + pq + p.\\n\\nNow, adding all these together:\\n\\npq + (pq + q\\u00b2 + q) + (p\\u00b2 + pq + p) = 191.\\n\\nCombine like terms:\\n\\npq + pq + pq = 3pq.\\n\\nq\\u00b2 + p\\u00b2.\\n\\nq + p.\\n\\nSo altogether, the equation becomes:\\n\\np\\u00b2 + q\\u00b2 + 3pq + p + q = 191.\\n\\nHmm. Let me note that.\\n\\nAlternatively, maybe I can factor this expression. Let me see. Let's try to group terms. Wait, another approach: since r = p + q + 1, then p + q + r = (p + q) + (p + q + 1) = 2(p + q) + 1. So if I can find p + q, then multiplying by 2 and adding 1 gives me the answer. That's useful. So perhaps instead of dealing with the first equation directly, I can express it in terms of p + q.\\n\\nBut maybe let's work with the equation we derived: p\\u00b2 + q\\u00b2 + 3pq + p + q = 191.\\n\\nWait, another thought: p\\u00b2 + q\\u00b2 + 3pq. Let's recall that (p + q)^2 = p\\u00b2 + 2pq + q\\u00b2, so p\\u00b2 + q\\u00b2 + 3pq = (p + q)^2 + pq.\\n\\nTherefore, the equation can be rewritten as:\\n\\n(p + q)^2 + pq + p + q = 191.\\n\\nLet me set S = p + q, and P = pq. Then the equation becomes:\\n\\nS\\u00b2 + P + S = 191.\\n\\nBut from the second equation, we know that r = S + 1, so since we need to find S + r = S + (S + 1) = 2S + 1. So our target is 2S + 1. So if we can find S, we can find the answer.\\n\\nNow, the equation is S\\u00b2 + P + S = 191. But S and P are related as S = p + q, P = pq. For two primes p and q, their sum and product. Since p and q are primes, maybe we can list possible primes that add up to S and multiply to P. However, since S and P are variables here, maybe we can express P in terms of S from the equation.\\n\\nFrom S\\u00b2 + P + S = 191, so P = 191 - S\\u00b2 - S.\\n\\nTherefore, P = -S\\u00b2 - S + 191. But P = pq must be positive, so -S\\u00b2 - S + 191 > 0. Therefore, S\\u00b2 + S < 191. Let's see the possible values of S. Since S is the sum of two primes, which are at least 2 each, so S is at least 2 + 2 = 4. Also, since S\\u00b2 + S < 191, let's solve S\\u00b2 + S - 191 < 0. Let's find the roots of S\\u00b2 + S - 191 = 0.\\n\\nUsing quadratic formula: S = [-1 \\u00b1 sqrt(1 + 4*191)] / 2 = [-1 \\u00b1 sqrt(765)] / 2. sqrt(765) is approx 27.66, so S \\u2248 (-1 + 27.66)/2 \\u2248 13.33. So the positive root is approximately 13.33, so S must be less than 13.33. Therefore, S can be integers from 4 up to 13.\\n\\nSo possible S values: 4,5,6,7,8,9,10,11,12,13.\\n\\nBut since p and q are primes, their sum S must be even or odd. Since except for 2, all primes are odd. So if both p and q are odd primes, their sum is even. If one is 2 and the other is odd, then their sum is odd. So S can be even or odd. So possible values of S (from 4 to 13) can be checked.\\n\\nBut maybe instead of all possible S, let's check possible S values from 4 to 13 and see if for each S, P = 191 - S\\u00b2 - S, and check if P can be expressed as the product of two primes that add up to S.\\n\\nAlternatively, maybe first check which S gives P as a product of two primes.\\n\\nLet's start with S=4. Then P = 191 - 16 -4 = 171. Then check if 171 can be written as product of two primes that add up to 4. But 4 is the sum. The primes could be 2 and 2 (since 2+2=4). Then 2*2=4, but P here is 171. 4 \\u2260 171, so S=4 is invalid.\\n\\nNext S=5: P=191 -25 -5=161. 161 factors into 7*23. Check if 7 + 23 = 30, which is not 5. So that's not possible. Alternatively, are there primes adding to 5? 2 and 3, since 2+3=5. Then P=2*3=6. But 6 \\u2260 161. So S=5 invalid.\\n\\nS=6: P=191 -36 -6=149. 149 is a prime number, so can't be expressed as product of two primes. So invalid.\\n\\nS=7: P=191 -49 -7=135. 135 factors into 5*27 (but 27 not prime), 3*45, 9*15, none primes. So no. So invalid.\\n\\nS=8: P=191 -64 -8=119. 119 factors into 7*17. Check if 7+17=24\\u22608. Primes adding to 8 are 3+5=8 or 5+3. Then P=15. But 15\\u2260119. So invalid.\\n\\nS=9: P=191 -81 -9=101. 101 is prime, so no.\\n\\nS=10: P=191 -100 -10=81. 81=9*9, but 9 not prime. So no.\\n\\nS=11: P=191 -121 -11=59. 59 is prime. So no.\\n\\nS=12: P=191 -144 -12=35. 35=5*7. Check if 5 +7=12? Yes! 5 +7=12. So here, S=12, which is p + q=12, and pq=35. 5 and 7 are primes. So this works.\\n\\nSo then p=5 and q=7, or p=7 and q=5. Then r = S + 1 =12 +1=13. Check if r is prime: 13 is prime. So yes. So then p, q, r are 5,7,13 or 7,5,13. Then p + q + r =5 +7 +13=25. So 25 would be the answer.\\n\\nWait, let me check S=13 as well just to be thorough. S=13: P=191 -169 -13=9. 9=3*3. Check if 3 +3=6\\u226013. So no. So no.\\n\\nTherefore, only S=12 gives valid primes. So the answer is 25.\\n\\nLet me verify the original equations. pq + qr + rp. Let p=5, q=7, r=13. Then 5*7 +7*13 +13*5 =35 +91 +65=35+91=126+65=191. Which matches. And p + q =5 +7=12. r -1=13 -1=12. So that also matches. So it's correct. Therefore, the answer is 25.\\n</think>To solve the problem where three prime numbers \\\\( p, q, \\\\) and \\\\( r \\\\) satisfy the equations \\\\( pq + qr + rp = 191 \\\\) and \\\\( p + q = r - 1 \\\\), we proceed as follows:\\n\\n1. **Express \\\\( r \\\\) in terms of \\\\( p \\\\) and \\\\( q \\\\):**\\n   From the equation \\\\( p + q = r - 1 \\\\), we can solve for \\\\( r \\\\):\\n   \\\\[\\n   r = p + q + 1\\n   \\\\]\\n\\n2. **Substitute \\\\( r \\\\) into the first equation:**\\n   Substitute \\\\( r = p + q + 1 \\\\) into the equation \\\\( pq + qr + rp = 191 \\\\):\\n   \\\\[\\n   pq + q(p + q + 1) + p(p + q + 1) = 191\\n   \\\\]\\n   Expanding and combining like terms:\\n   \\\\[\\n   pq + pq + q^2 + q + p^2 + pq + p = 191\\n   \\\\]\\n   Simplify:\\n   \\\\[\\n   p^2 + q^2 + 3pq + p + q = 191\\n   \\\\]\\n\\n3. **Introduce new variables:**\\n   Let \\\\( S = p + q \\\\) and \\\\( P = pq \\\\). The equation becomes:\\n   \\\\[\\n   S^2 + P + S = 191\\n   \\\\]\\n\\n4. **Express \\\\( r \\\\) in terms of \\\\( S \\\\):**\\n   Since \\\\( r = p + q + 1 = S + 1 \\\\), we need to find \\\\( S \\\\) such that \\\\( S^2 + P + S = 191 \\\\) and \\\\( P = pq \\\\) is the product of two primes \\\\( p \\\\) and \\\\( q \\\\) that sum to \\\\( S \\\\).\\n\\n5. **Determine possible values for \\\\( S \\\\):**\\n   Solve the inequality \\\\( S^2 + S < 191 \\\\):\\n   \\\\[\\n   S^2 + S - 191 < 0\\n   \\\\]\\n   Using the quadratic formula \\\\( S = \\\\frac{-1 \\\\pm \\\\sqrt{1 + 4 \\\\cdot 191}}{2} \\\\):\\n   \\\\[\\n   S = \\\\frac{-1 \\\\pm \\\\sqrt{765}}{2}\\n   \\\\]\\n   Since \\\\( \\\\sqrt{765} \\\\approx 27.66 \\\\), we have:\\n   \\\\[\\n   S \\\\approx \\\\frac{-1 + 27.66}{2} \\\\approx 13.33\\n   \\\\]\\n   Therefore, \\\\( S \\\\) must be an integer between 4 and 13.\\n\\n6. **Check possible values of \\\\( S \\\\):**\\n   - For \\\\( S = 12 \\\\):\\n     \\\\[\\n     P = 191 - 12^2 - 12 = 191 - 144 - 12 = 35\\n     \\\\]\\n     Check if \\\\( 35 \\\\) can be written as the product of two primes that sum to 12:\\n     \\\\[\\n     35 = 5 \\\\times 7 \\\\quad \\\\text{and} \\\\quad 5 + 7 = 12\\n     \\\\]\\n     This works. So \\\\( p = 5 \\\\) and \\\\( q = 7 \\\\).\\n\\n7. **Find \\\\( r \\\\):**\\n   \\\\[\\n   r = S + 1 = 12 + 1 = 13\\n   \\\\]\\n\\n8. **Verify the solution:**\\n   - Check \\\\( pq + qr + rp = 191 \\\\):\\n     \\\\[\\n     5 \\\\times 7 + 7 \\\\times 13 + 13 \\\\times 5 = 35 + 91 + 65 = 191\\n     \\\\]\\n   - Check \\\\( p + q = r - 1 \\\\):\\n     \\\\[\\n     5 + 7 = 12 \\\\quad \\\\text{and} \\\\quad 13 - 1 = 12\\n     \\\\]\\n\\nSince all conditions are satisfied, the final answer is:\\n\\\\[\\n\\\\boxed{25}\\n\\\\]\",\n          \"<think>\\nOkay, let's see. I need to solve this problem where x and y are positive integers satisfying 2(x + y) = gcd(x, y) + lcm(x, y). And I have to find the ratio of the lcm to the gcd of x and y. Hmm, okay, let's break this down.\\n\\nFirst, I remember that for any two positive integers, the product of the lcm and gcd of those numbers is equal to the product of the numbers themselves. So, lcm(x, y) * gcd(x, y) = x * y. That might come in handy here. Let me note that down: lcm(x,y)*gcd(x,y) = x*y.\\n\\nGiven the equation 2(x + y) = gcd(x, y) + lcm(x, y), maybe I can express everything in terms of gcd and the ratio of x and y. Since gcd and lcm are involved, it might be helpful to let d = gcd(x, y), and then express x and y as x = d*a and y = d*b, where a and b are coprime integers (their gcd is 1). That's a standard approach for problems involving gcd and lcm.\\n\\nSo let me set d = gcd(x, y). Then x = d*a, y = d*b, with gcd(a, b) = 1. Then, the lcm(x, y) would be d*a*b, because lcm(x, y) = x*y / gcd(x, y) = (d*a*d*b)/d = d*a*b. Right, so lcm(x,y) = d*a*b.\\n\\nSubstituting these into the original equation: 2(x + y) = gcd(x, y) + lcm(x, y).\\n\\nSubstituting x = d*a, y = d*b, gcd = d, lcm = d*a*b. Then:\\n\\n2(d*a + d*b) = d + d*a*b.\\n\\nFactor out d from the left side: 2d(a + b) = d(1 + a*b).\\n\\nSince d is a positive integer, we can divide both sides by d, yielding:\\n\\n2(a + b) = 1 + a*b.\\n\\nSo now the equation simplifies to 2(a + b) = a*b + 1, where a and b are coprime positive integers. Hmm, okay. Now we have a simpler equation to solve: a*b - 2a - 2b + 1 = 0. Let me rearrange that:\\n\\na*b - 2a - 2b + 1 = 0.\\n\\nHmm, maybe factor this equation? Let me see. Adding 4 to both sides might help in factoring. Let's try:\\n\\na*b - 2a - 2b + 1 + 4 - 4 = 0\\n\\nSo, a*b - 2a - 2b + 4 = 3.\\n\\nWait, not sure. Alternatively, perhaps rearrange the terms:\\n\\na*b - 2a - 2b = -1.\\n\\nThen, add 4 to both sides:\\n\\na*b - 2a - 2b + 4 = 3.\\n\\nNow, left side can be factored as (a - 2)(b - 2) = 3. Because expanding (a - 2)(b - 2) gives a*b - 2a - 2b + 4. Yes, that's right. So:\\n\\n(a - 2)(b - 2) = 3.\\n\\nSince a and b are positive integers and coprime, we need to find pairs (a, b) such that their product is 3 when each is reduced by 2. Also, since a and b are coprime, (a - 2) and (b - 2) must be divisors of 3, which is prime. The positive divisors of 3 are 1 and 3.\\n\\nSo possible pairs (since a and b are positive integers, a - 2 and b - 2 must be at least such that a and b are positive. Let's see:\\n\\nCase 1: (a - 2) = 1 and (b - 2) = 3. Then, a = 3, b = 5. Check if gcd(a, b) = 1. gcd(3, 5) = 1, which is good.\\n\\nCase 2: (a - 2) = 3 and (b - 2) = 1. Then, a = 5, b = 3. Similarly, gcd(5, 3) = 1. So this is also valid.\\n\\nBut also, since 3 is prime, the only positive divisors are 1 and 3. But since we're considering positive integers, we could also consider if one of them is negative? But since a and b are positive, a - 2 and b - 2 must be positive or zero? Wait, but 3 is positive, so the factors must both be positive. Because if one of (a - 2) or (b - 2) were negative, their product would be negative, but 3 is positive. So both (a - 2) and (b - 2) must be positive. Thus, only the two cases above.\\n\\nAlternatively, maybe (a - 2) and (b - 2) could be 3 and 1 in some order, which gives the two cases. So the possible (a, b) are (3, 5) and (5, 3). Since a and b are interchangeable (since x and y are symmetric in the problem), these two cases would yield the same results.\\n\\nSo now, let's see. For (a, b) = (3, 5), then x = d*3, y = d*5. Similarly, for (a, b) = (5, 3), x = d*5, y = d*3. But since the problem is symmetric in x and y, both cases are equivalent.\\n\\nNow, since we need to find the ratio lcm(x, y)/gcd(x, y), let's compute that.\\n\\nRecall that lcm(x, y)/gcd(x, y) = (d*a*b)/d = a*b. So it's simply a*b. Since in both cases, a and b are 3 and 5, the product is 15. Therefore, the ratio is 15.\\n\\nWait, that seems too straightforward. Let me check.\\n\\nIf the ratio is a*b, then yes. Because lcm(x, y) is d*a*b and gcd(x, y) is d, so their ratio is (d*a*b)/d = a*b. Since a and b are 3 and 5, 3*5=15. Therefore, the answer is 15. So the answer is 15.\\n\\nBut let me verify with an example. Let's take d=1. Then x=3, y=5. Then gcd(3,5)=1, lcm=15. Then 2(x + y) = 2*(8) = 16. The right side is 1 + 15=16. So that works. If d=1, then 2(3+5)=16=1 +15.\\n\\nWhat if d=2? Then x=6, y=10. gcd(6,10)=2, lcm=30. Then 2(6 + 10)=2*16=32. The right side is 2 +30=32. So that also works. Then the ratio lcm/gcd is 30/2=15. So regardless of d, the ratio is always 15. Wait, because if x = d*a, y = d*b, then lcm(x,y)/gcd(x,y) = (d*a*b)/d = a*b, which is 15 as in the first case. So regardless of d, the ratio is always 15. Therefore, the answer is 15.\\n\\nSo even if d is some other positive integer, the ratio remains a*b =15. Therefore, the required ratio is 15.\\n\\nTherefore, the answer is 15. So \\\\boxed{15}.\\n\\n**Final Answer**\\n\\\\boxed{15}\\n</think>Given \\\\( x \\\\) and \\\\( y \\\\) are positive integers such that \\\\( 2(x + y) = \\\\gcd(x, y) + \\\\text{lcm}(x, y) \\\\), we need to find \\\\( \\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} \\\\).\\n\\nFirst, let \\\\( d = \\\\gcd(x, y) \\\\). Then, we can express \\\\( x \\\\) and \\\\( y \\\\) as \\\\( x = d \\\\cdot a \\\\) and \\\\( y = d \\\\cdot b \\\\), where \\\\( \\\\gcd(a, b) = 1 \\\\). The least common multiple (lcm) of \\\\( x \\\\) and \\\\( y \\\\) is given by \\\\( \\\\text{lcm}(x, y) = d \\\\cdot a \\\\cdot b \\\\).\\n\\nSubstituting these into the given equation:\\n\\\\[\\n2(d \\\\cdot a + d \\\\cdot b) = d + d \\\\cdot a \\\\cdot b\\n\\\\]\\nDividing both sides by \\\\( d \\\\):\\n\\\\[\\n2(a + b) = 1 + a \\\\cdot b\\n\\\\]\\nRearranging terms, we get:\\n\\\\[\\na \\\\cdot b - 2a - 2b + 1 = 0\\n\\\\]\\nAdding 4 to both sides to factorize:\\n\\\\[\\na \\\\cdot b - 2a - 2b + 4 = 3\\n\\\\]\\nThis can be factored as:\\n\\\\[\\n(a - 2)(b - 2) = 3\\n\\\\]\\nThe positive integer solutions for \\\\((a - 2)\\\\) and \\\\((b - 2)\\\\) are 1 and 3. Thus, the possible pairs \\\\((a, b)\\\\) are \\\\((3, 5)\\\\) and \\\\((5, 3)\\\\). Since \\\\( a \\\\) and \\\\( b \\\\) are coprime, both pairs are valid.\\n\\nThe ratio \\\\( \\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} \\\\) is given by:\\n\\\\[\\n\\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} = \\\\frac{d \\\\cdot a \\\\cdot b}{d} = a \\\\cdot b\\n\\\\]\\nFor both pairs \\\\((3, 5)\\\\) and \\\\((5, 3)\\\\), the product \\\\( a \\\\cdot b = 15 \\\\).\\n\\nThus, the final answer is:\\n\\\\[\\n\\\\boxed{15}\\n\\\\]\",\n          \"<think>\\nOkay, so I need to find the remainder when the product of all odd numbers from 1 to 2005 is divided by 1000. Hmm, let's think about how to approach this. \\n\\nFirst, the product is 1 \\u00d7 3 \\u00d7 5 \\u00d7 ... \\u00d7 2005. That's a lot of numbers! Since we're dealing with division by 1000, maybe modular arithmetic can help here. The remainder when divided by 1000 is equivalent to the product modulo 1000. But calculating such a huge product directly seems impossible. There has to be a smarter way.\\n\\nI remember that when dealing with factorials and remainders, factors of 2 and 5 can create trailing zeros. However, here we're only multiplying odd numbers, so there are no factors of 2. But there might still be factors of 5. Wait, 1000 is 8\\u00d7125, which is 2^3 \\u00d7 5^3. Since the product is all odd numbers, it won't have factors of 2, but it can have factors of 5. Therefore, the product will be divisible by 5^3, but since there are no 2s, the product modulo 1000 might not be zero. Hmm, maybe I need to compute the product modulo 1000, but adjusting for the factors of 5?\\n\\nAlternatively, maybe split the problem into modulo 8 and modulo 125, then use the Chinese Remainder Theorem (CRT) to combine the results. Since 1000 = 8 \\u00d7 125, and 8 and 125 are coprime, CRT says that if I can find the remainder modulo 8 and modulo 125, then I can combine them to find the remainder modulo 1000. That might be a good approach.\\n\\nLet me start with modulo 8. The product is 1\\u00d73\\u00d75\\u00d77\\u00d79\\u00d7...\\u00d72005. But modulo 8, odd numbers repeat every 8 numbers. Let's see, the residues modulo 8 of odd numbers are 1,3,5,7,1,3,5,7,... So the pattern repeats every 4 terms. Wait, no. Wait, the numbers go 1,3,5,7,9\\u22611,11\\u22613,13\\u22615,15\\u22617, etc. So every 8 numbers, the cycle of residues 1,3,5,7 repeats twice. Wait, actually, modulo 8, the odd residues cycle every 4 numbers. Let's confirm:\\n\\n1 mod 8 =1\\n\\n3 mod8=3\\n\\n5 mod8=5\\n\\n7 mod8=7\\n\\n9 mod8=1\\n\\n11 mod8=3\\n\\n13 mod8=5\\n\\n15 mod8=7\\n\\nYes, every 4 terms, the cycle repeats. So how many terms are in the product 1\\u00d73\\u00d75\\u00d7...\\u00d72005? Let's find the number of terms first. The nth odd number is 2n-1. So 2n-1=2005 => n=(2005+1)/2=2006/2=1003. So there are 1003 terms.\\n\\nSo 1003 terms, each group of 4 terms (mod8) is 1\\u00d73\\u00d75\\u00d77=105. Then 105 mod8= 105 - 13\\u00d78=105-104=1. So each group of 4 terms multiplies to 1 mod8. Then how many full groups of 4 are there in 1003 terms? Let's divide 1003 by 4. 1003 \\u00f74=250.75. So 250 full groups, each contributing 1 mod8, and then a remainder of 3 terms. \\n\\nSo the total product modulo8 is (1^250) \\u00d7 (last three terms). The last three terms would be the terms after the 250th group. The 250th group ends at term 250\\u00d74=1000. So the 1001st term is 2\\u00d71001 -1=2001. Wait, no: the first term is 1=2\\u00d71-1, second term 3=2\\u00d72-1, so term k is 2k-1. Therefore, term 1001 is 2\\u00d71001 -1=2002-1=2001. Then the 1002nd term is 2003, 1003rd term is 2005. So the last three terms are 2001, 2003, 2005. Let's compute each mod8:\\n\\n2001 \\u00f78: 8\\u00d7250=2000, so 2001 mod8=1\\n\\n2003 mod8=3\\n\\n2005 mod8=5\\n\\nSo the last three terms modulo8 are 1\\u00d73\\u00d75=15 mod8=7.\\n\\nTherefore, total product mod8 is (1^250) \\u00d77=1\\u00d77=7 mod8.\\n\\nSo the remainder modulo8 is7.\\n\\nNow, we need to compute the product modulo125. This seems more complicated. Let's think.\\n\\nThe product is the product of all odd numbers from1 to2005. Wait, 2005=5\\u00d7401. So we can write the product as (1\\u00d73\\u00d75\\u00d77\\u00d7...\\u00d72005). Let's note that there are a lot of factors of 5 in this product, which would make the product divisible by 5 multiple times. However, modulo125 is 5^3, so if the product has at least three factors of 5, then modulo125 would be 0. Wait, but maybe even if it's divisible by 5^3, but we need to compute the actual remainder. Wait, but perhaps the product is divisible by 5^3, but when divided by 5^3, the remaining product modulo8 or something else. Wait, maybe not. Let me check how many factors of5 are in the product.\\n\\nThe number of factors of5 in the product:\\n\\nEach multiple of5 contributes at least one factor of5. Since we're dealing with odd numbers, the multiples of5 that are odd. So numbers divisible by5 but not by2. So numbers like5,15,25,...,2005. Let's count how many multiples of5 are in the product. The first term is5, which is5\\u00d71, then15=5\\u00d73,..., up to2005=5\\u00d7401. So the multiples of5 are5\\u00d7(1,3,5,...,401). Wait, 5\\u00d7k, where k is odd from1 to401. Because 5\\u00d7401=2005. So how many terms are there?\\n\\nThe number of terms k from1 to401 where k is odd. Since401 is odd, the number is (401 +1)/2=201. So there are201 multiples of5 in the product. Each contributes at least one factor of5. Additionally, multiples of25 contribute an extra factor of5. Similarly, multiples of125, 625, etc., contribute more factors.\\n\\nSo let's compute the total number of factors of5 in the product.\\n\\nNumber of multiples of5:201 (as above)\\n\\nNumber of multiples of25: These are numbers in the product divisible by25. Since the product includes numbers of the form5\\u00d7(odd numbers). So multiples of25 are numbers divisible by25, which are 25,75,125,...,2000. But 2005 is not divisible by25. Wait, wait, in the original product (all odd numbers up to2005), the multiples of25 must be odd multiples. So 25\\u00d71,25\\u00d73,..., up to the largest odd multiple less than or equal to2005.\\n\\n25\\u00d7k \\u22642005, where k is odd. Let's compute k_max:\\n\\n25k \\u22642005 => k \\u22642005/25=80.2. So the largest integer k is80, but since k has to be odd, the largest odd k is79. So 25\\u00d779=1975. Then 25\\u00d781=2025, which is over. So the multiples of25 in the product are25\\u00d71,25\\u00d73,...,25\\u00d779. Number of terms: (79-1)/2 +1=39 +1=40. Wait, from1 to79 odd numbers: number is (79+1)/2=40. So there are40 multiples of25.\\n\\nSimilarly, multiples of125:125\\u00d71,125\\u00d73,..., up to125\\u00d7k\\u22642005. 125\\u00d7k \\u22642005 =>k\\u226416.04. So k_max=15 (odd). So 125\\u00d715=1875. So multiples are125\\u00d71,125\\u00d73,...,125\\u00d715. Number of terms: (15-1)/2 +1=7+1=8.\\n\\nMultiples of625:625\\u00d71=625, next is625\\u00d73=1875, next is625\\u00d75=3125>2005. So only two multiples:625 and1875. But1875 is already counted as a multiple of125. So factors of625 contribute an extra factor of5 each. So number of multiples of625 is 2. 625 and1875.\\n\\nMultiples of3125:3125>2005, so none.\\n\\nSo total number of factors of5:\\n\\nFrom multiples of5:201\\n\\nFrom multiples of25:40 (each contributes an extra)\\n\\nFrom multiples of125:8 (each contributes another extra)\\n\\nFrom multiples of625:2 (each contributes another extra)\\n\\nTotal:201 +40 +8 +2=251.\\n\\nSo total factors of5 in the product:251.\\n\\nSimilarly, factors of2: since all numbers are odd, there are none. So the product is divisible by5^251 but not by2. So when dividing by5^3, since 251\\u22653, the product is divisible by5^3. Therefore, the product modulo125 is0? Wait, no, wait. Wait, 125 is5^3. If the product has at least three factors of5, then when divided by5^3, the quotient is an integer, but the remainder when divided by125 is0. So if the product is divisible by125, then the remainder is0. But the question is, when we divide the product by1000, which is8\\u00d7125, the remainder is to be found. However, we already considered that modulo8 is7, and modulo125 is... Hmm, but if modulo125 is0, then using CRT, we can say the remainder is a number congruent to7 mod8 and0 mod125. So we need to solve for x\\u22610 mod125 andx\\u22617 mod8.\\n\\nBut wait, let me confirm if modulo125 is indeed0. Let's check: since the product has at least three factors of5, then yes, the product is divisible by5^3, so product \\u22610 mod125. Therefore, modulo125 is0. Therefore, we have:\\n\\nx \\u22617 mod8\\n\\nx \\u22610 mod125\\n\\nWe need to find x such that x \\u22610 mod125 andx \\u22617 mod8. Let\\u2019s solve this system.\\n\\nLet x=125k. Then 125k\\u22617 mod8. Since125 mod8=5, so 5k\\u22617 mod8.\\n\\nWe solve 5k\\u22617 mod8.\\n\\nMultiply both sides by inverse of5 mod8. The inverse of5 mod8 is5, since5\\u00d75=25\\u22611 mod8. So multiply both sides by5:\\n\\nk\\u22617\\u00d75 mod8 =>k\\u226135 mod8 =>35\\u00f78=4*8=32, 35-32=3. So k\\u22613 mod8.\\n\\nTherefore, k=8m +3 for some integer m. Therefore, x=125(8m +3)=1000m +375. Therefore, the smallest non-negative solution is375. Therefore, the remainder is375 when divided by1000. So the answer is375.\\n\\nWait, but before accepting that, let me verify my steps again because this is tricky.\\n\\nFirst, confirming that the product is divisible by5^3: yes, since there are 251 factors of5, so 5^251 divides the product. Therefore, the product is divisible by5^3, so product\\u22610 mod125. That's correct.\\n\\nThen, solving x\\u22617 mod8 and x\\u22610 mod125. So x=125k. Then 125k\\u22615k mod8. So 5k\\u22617 mod8. Multiply both sides by inverse of5 mod8, which is5, since5\\u00d75=25\\u22611 mod8. So k\\u226135 mod8\\u22613 mod8. Therefore, k=8m +3, so x=125\\u00d73 +1000m=375 +1000m. So the minimal positive solution is375. So remainder is375 when divided by1000. That seems correct.\\n\\nBut wait, to make sure, let me check with an example. Let's take x=375. 375 \\u00f78=46*8=368, remainder7. So 375 mod8=7. 375 mod125=0. So yes, 375 satisfies both conditions. Therefore, the remainder is375. Therefore, the answer is\\\\boxed{375}.\\n\\nWait, but I need to make sure that my calculation for modulo8 and modulo125 are correct. Let me double-check the modulo8 calculation.\\n\\nEarlier, I considered that the product modulo8 is7. Let me recast that.\\n\\nThe product is1\\u00d73\\u00d75\\u00d77\\u00d7\\u2026\\u00d72005. When taking modulo8, each cycle of4 terms (1,3,5,7) multiplies to1\\u00d73\\u00d75\\u00d77=105\\u22611 mod8. Since 105/8=13\\u00d78=104, 105-104=1. Then, since there are1003 terms, how many cycles of4 are there? 1003 divided by4 is250 cycles with a remainder of3 terms. Then the product is (1)^250 \\u00d7 last3 terms.\\n\\nThe last3 terms are2001,2003,2005. 2001 mod8=1 (2001-8\\u00d7250=2001-2000=1), 2003=2001+2\\u21921+2=3 mod8, 2005=2001+4\\u21921+4=5 mod8. So last3 terms are1\\u00d73\\u00d75=15\\u22617 mod8. Therefore, total product\\u22611^250 \\u00d77\\u22617 mod8. Correct.\\n\\nTherefore, the logic holds. Therefore, the remainder is375. So I think that's the correct answer.\\n\\n**Final Answer**\\n\\\\boxed{375}\\n</think>To find the remainder when the product \\\\(1 \\\\times 3 \\\\times 5 \\\\times \\\\cdots \\\\times 2005\\\\) is divided by 1000, we use modular arithmetic and the Chinese Remainder Theorem (CRT).\\n\\n### Step 1: Calculate the product modulo 8\\n\\nThe sequence of odd numbers modulo 8 repeats every 4 terms: \\\\(1, 3, 5, 7\\\\). The product of each cycle is:\\n\\\\[\\n1 \\\\times 3 \\\\times 5 \\\\times 7 = 105 \\\\equiv 1 \\\\mod 8\\n\\\\]\\n\\nThere are 1003 terms in the product. Dividing 1003 by 4 gives 250 full cycles and a remainder of 3 terms. The remaining terms are 2001, 2003, and 2005. We calculate these modulo 8:\\n\\\\[\\n2001 \\\\equiv 1 \\\\mod 8, \\\\quad 2003 \\\\equiv 3 \\\\mod 8, \\\\quad 2005 \\\\equiv 5 \\\\mod 8\\n\\\\]\\nThe product of these remaining terms is:\\n\\\\[\\n1 \\\\times 3 \\\\times 5 = 15 \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\nThus, the product modulo 8 is:\\n\\\\[\\n1^{250} \\\\times 7 \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\n### Step 2: Calculate the product modulo 125\\n\\nWe need to determine the number of factors of 5 in the product. The sequence of odd numbers includes multiples of 5, 25, 125, and 625.\\n\\n- Multiples of 5: \\\\(5, 15, 25, \\\\ldots, 2005\\\\)\\n  - These are of the form \\\\(5 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 200\\\\)\\n  - Number of such terms: \\\\(\\\\frac{2005}{5} = 401\\\\), and half of these are odd, so \\\\(201\\\\) multiples of 5.\\n\\n- Multiples of 25: \\\\(25, 75, 125, \\\\ldots, 1975\\\\)\\n  - These are of the form \\\\(25 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 39\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1975}{25} = 79\\\\), and half of these are odd, so \\\\(40\\\\) multiples of 25.\\n\\n- Multiples of 125: \\\\(125, 375, 625, 875, 1125, 1375, 1625, 1875\\\\)\\n  - These are of the form \\\\(125 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 15\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1875}{125} = 15\\\\), and half of these are odd, so \\\\(8\\\\) multiples of 125.\\n\\n- Multiples of 625: \\\\(625, 1875\\\\)\\n  - These are of the form \\\\(625 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1875}{625} = 3\\\\), and half of these are odd, so \\\\(2\\\\) multiples of 625.\\n\\nTotal factors of 5:\\n\\\\[\\n201 + 40 + 8 + 2 = 251\\n\\\\]\\n\\nSince \\\\(251 \\\\geq 3\\\\), the product is divisible by \\\\(5^3 = 125\\\\). Therefore, the product modulo 125 is:\\n\\\\[\\n0 \\\\mod 125\\n\\\\]\\n\\n### Step 3: Combine results using the Chinese Remainder Theorem\\n\\nWe have:\\n\\\\[\\nx \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\\\[\\nx \\\\equiv 0 \\\\mod 125\\n\\\\]\\n\\nLet \\\\(x = 125k\\\\). Then:\\n\\\\[\\n125k \\\\equiv 7 \\\\mod 8\\n\\\\]\\nSince \\\\(125 \\\\equiv 5 \\\\mod 8\\\\), we have:\\n\\\\[\\n5k \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\nThe multiplicative inverse of 5 modulo 8 is 5, so:\\n\\\\[\\nk \\\\equiv 7 \\\\times 5 \\\\equiv 35 \\\\equiv 3 \\\\mod 8\\n\\\\]\\n\\nThus, \\\\(k = 8m + 3\\\\) for some integer \\\\(m\\\\). Therefore:\\n\\\\[\\nx = 125(8m + 3) = 1000m + 375\\n\\\\]\\n\\nThe smallest non-negative solution is:\\n\\\\[\\nx = 375\\n\\\\]\\n\\nTherefore, the remainder when the product is divided by 1000 is:\\n\\\\[\\n\\\\boxed{375}\\n\\\\]\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    }\n  ]\n}",
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       "    <path d=\"M7,19H8.4L18.45,9,17,7.55,7,17.6ZM5,21V16.75L18.45,3.32a2,2,0,0,1,2.83,0l1.4,1.43a1.91,1.91,0,0,1,.58,1.4,1.91,1.91,0,0,1-.58,1.4L9.25,21ZM18.45,9,17,7.55Zm-12,3A5.31,5.31,0,0,0,4.9,8.1,5.31,5.31,0,0,0,1,6.5,5.31,5.31,0,0,0,4.9,4.9,5.31,5.31,0,0,0,6.5,1,5.31,5.31,0,0,0,8.1,4.9,5.31,5.31,0,0,0,12,6.5,5.46,5.46,0,0,0,6.5,12Z\"/>\n",
       "  </svg>\n",
       "    </button>\n",
       "    <script>\n",
       "      (() => {\n",
       "      const buttonEl =\n",
       "        document.querySelector('#id_1f2d874d-d63f-4045-815d-f673a3a4fa5e button.colab-df-generate');\n",
       "      buttonEl.style.display =\n",
       "        google.colab.kernel.accessAllowed ? 'block' : 'none';\n",
       "\n",
       "      buttonEl.onclick = () => {\n",
       "        google.colab.notebook.generateWithVariable('dataset');\n",
       "      }\n",
       "      })();\n",
       "    </script>\n",
       "  </div>\n",
       "\n",
       "    </div>\n",
       "  </div>\n"
      ],
      "text/plain": [
       "      expected_answer                                            problem  \\\n",
       "0                  14  Given $\\sqrt{x^2+165}-\\sqrt{x^2-52}=7$ and $x$...   \n",
       "6                  -2  Find the value of the parameter $a$ for which ...   \n",
       "9                  18  What is the sum of all real numbers $x$ for wh...   \n",
       "13                  2  Evaluate the sum \\(\\sum_{n=1}^\\infty \\frac{\\ph...   \n",
       "17                 30  What is the largest positive integer that divi...   \n",
       "...               ...                                                ...   \n",
       "19243             244  Let \\( p \\), \\( q \\), and \\( r \\) be the disti...   \n",
       "19245               1  A bug is on the $0$ of a number line. At any p...   \n",
       "19247               4  A bus left point X for point Y. Two hours late...   \n",
       "19248              18  Each interior angle of a regular n-gon measure...   \n",
       "19250          0.8960  Find the probability that the second blue resu...   \n",
       "\n",
       "                                      generated_solution  \n",
       "0      <think>\\nOkay, let's see. I need to solve the ...  \n",
       "6      <think>\\nOkay, so I need to find the value of ...  \n",
       "9      <think>\\nOkay, so I need to solve the equation...  \n",
       "13     <think>\\nOkay, so I need to evaluate the infin...  \n",
       "17     <think>\\nAlright, so I need to find the larges...  \n",
       "...                                                  ...  \n",
       "19243  <think>\\nOkay, so I need to find the value of ...  \n",
       "19245  <think>\\nOkay, so I have this problem where a ...  \n",
       "19247  <think>\\nOkay, let's tackle this problem step ...  \n",
       "19248  <think>\\nOkay, let's see. I need to find the n...  \n",
       "19250  <think>\\nOkay, so I need to find the probabili...  \n",
       "\n",
       "[7507 rows x 3 columns]"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from datasets import load_dataset\n",
    "import pandas as pd\n",
    "import numpy as np\n",
    "\n",
    "dataset = load_dataset(\"unsloth/OpenMathReasoning-mini\", split = \"cot\")\n",
    "dataset = dataset.to_pandas()[\n",
    "    [\"expected_answer\", \"problem\", \"generated_solution\"]\n",
    "]\n",
    "\n",
    "# Try converting to number - if not, replace with NaN\n",
    "is_number = pd.to_numeric(pd.Series(dataset[\"expected_answer\"]), errors = \"coerce\").notnull()\n",
    "# Select only numbers\n",
    "dataset = dataset.iloc[np.where(is_number)[0]]\n",
    "\n",
    "dataset"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "JVRFqoSdIEVK"
   },
   "source": [
    "We have to format the dataset to follow our GRPO style formatting:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "id": "Z9ydcV_Abfi6"
   },
   "outputs": [],
   "source": [
    "def format_dataset(x):\n",
    "    expected_answer = x[\"expected_answer\"]\n",
    "    problem = x[\"problem\"]\n",
    "\n",
    "    # Remove generated <think> and </think>\n",
    "    thoughts = x[\"generated_solution\"]\n",
    "    thoughts = thoughts.replace(\"<think>\", \"\").replace(\"</think>\", \"\")\n",
    "\n",
    "    # Strip newlines on left and right\n",
    "    thoughts = thoughts.strip()\n",
    "    # Add our custom formatting\n",
    "    final_prompt = \\\n",
    "        reasoning_start + thoughts + reasoning_end + \\\n",
    "        solution_start + expected_answer + solution_end\n",
    "    return [\n",
    "        {\"role\" : \"system\",    \"content\" : system_prompt},\n",
    "        {\"role\" : \"user\",      \"content\" : problem},\n",
    "        {\"role\" : \"assistant\", \"content\" : final_prompt},\n",
    "    ]\n",
    "\n",
    "dataset[\"Messages\"] = dataset.apply(format_dataset, axis = 1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "X5NI47rOIRP2"
   },
   "source": [
    "Check to see if it worked:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 140
    },
    "id": "LTdXBKcslhRH",
    "outputId": "2803eb03-4185-4f1b-a6fe-5d4b2b4f0f56"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "\"You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION><|eot_id|>Given $\\\\sqrt{x^2+165}-\\\\sqrt{x^2-52}=7$ and $x$ is positive, find all possible values of $x$.<start_working_out>Okay, let's see. I need to solve the equation \u221a(x\u00b2 + 165) - \u221a(x\u00b2 - 52) = 7, and find all positive values of x. Hmm, radicals can be tricky, but maybe if I can eliminate the square roots by squaring both sides. Let me try that.\\n\\nFirst, let me write down the equation again to make sure I have it right:\\n\\n\u221a(x\u00b2 + 165) - \u221a(x\u00b2 - 52) = 7.\\n\\nOkay, so the idea is to isolate one of the radicals and then square both sides. Let me try moving the second radical to the other side:\\n\\n\u221a(x\u00b2 + 165) = 7 + \u221a(x\u00b2 - 52).\\n\\nNow, if I square both sides, maybe I can get rid of the square roots. Let's do that:\\n\\n(\u221a(x\u00b2 + 165))\u00b2 = (7 + \u221a(x\u00b2 - 52))\u00b2.\\n\\nSimplifying the left side:\\n\\nx\u00b2 + 165 = 49 + 14\u221a(x\u00b2 - 52) + (\u221a(x\u00b2 - 52))\u00b2.\\n\\nThe right side is expanded using the formula (a + b)\u00b2 = a\u00b2 + 2ab + b\u00b2. So the right side becomes 7\u00b2 + 2*7*\u221a(x\u00b2 - 52) + (\u221a(x\u00b2 - 52))\u00b2, which is 49 + 14\u221a(x\u00b2 - 52) + (x\u00b2 - 52).\\n\\nSo putting it all together:\\n\\nx\u00b2 + 165 = 49 + 14\u221a(x\u00b2 - 52) + x\u00b2 - 52.\\n\\nHmm, let's simplify the right side. The x\u00b2 terms will cancel out, right? Let's subtract x\u00b2 from both sides:\\n\\n165 = 49 + 14\u221a(x\u00b2 - 52) - 52.\\n\\nSimplify the constants on the right:\\n\\n49 - 52 is -3, so:\\n\\n165 = -3 + 14\u221a(x\u00b2 - 52).\\n\\nNow, add 3 to both sides to isolate the radical term:\\n\\n165 + 3 = 14\u221a(x\u00b2 - 52).\\n\\nSo 168 = 14\u221a(x\u00b2 - 52).\\n\\nDivide both sides by 14:\\n\\n168 / 14 = \u221a(x\u00b2 - 52).\\n\\n12 = \u221a(x\u00b2 - 52).\\n\\nNow, square both sides again to eliminate the square root:\\n\\n12\u00b2 = x\u00b2 - 52.\\n\\n144 = x\u00b2 - 52.\\n\\nAdd 52 to both sides:\\n\\n144 + 52 = x\u00b2.\\n\\n196 = x\u00b2.\\n\\nSo x = \u221a196 = 14.\\n\\nBut wait, since the problem states that x is positive, we only take the positive root. So x = 14.\\n\\nBut hold on, when dealing with squaring equations, sometimes extraneous solutions can come up. I should check if this solution actually satisfies the original equation.\\n\\nLet's plug x = 14 back into the original equation:\\n\\n\u221a(14\u00b2 + 165) - \u221a(14\u00b2 - 52) = ?\\n\\nCalculate each term:\\n\\n14\u00b2 is 196.\\n\\nSo first radical: \u221a(196 + 165) = \u221a361 = 19.\\n\\nSecond radical: \u221a(196 - 52) = \u221a144 = 12.\\n\\nSo 19 - 12 = 7, which is exactly the right-hand side. So yes, it checks out.\\n\\nTherefore, the only solution is x = 14. Since the problem says x is positive, we don't have to consider negative roots. So I think that's the answer.\\nTo solve the equation \\\\(\\\\sqrt{x^2 + 165} - \\\\sqrt{x^2 - 52} = 7\\\\) for positive \\\\(x\\\\), we proceed as follows:\\n\\n1. Start with the given equation:\\n   \\\\[\\n   \\\\sqrt{x^2 + 165} - \\\\sqrt{x^2 - 52} = 7\\n   \\\\]\\n\\n2. Isolate one of the square roots by moving \\\\(\\\\sqrt{x^2 - 52}\\\\) to the right side:\\n   \\\\[\\n   \\\\sqrt{x^2 + 165} = 7 + \\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n3. Square both sides to eliminate the square root on the left:\\n   \\\\[\\n   (\\\\sqrt{x^2 + 165})^2 = (7 + \\\\sqrt{x^2 - 52})^2\\n   \\\\]\\n   Simplifying both sides, we get:\\n   \\\\[\\n   x^2 + 165 = 49 + 14\\\\sqrt{x^2 - 52} + (x^2 - 52)\\n   \\\\]\\n\\n4. Combine like terms on the right side:\\n   \\\\[\\n   x^2 + 165 = x^2 - 52 + 49 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n   Simplifying further:\\n   \\\\[\\n   x^2 + 165 = x^2 - 3 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n5. Subtract \\\\(x^2\\\\) from both sides:\\n   \\\\[\\n   165 = -3 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n6. Add 3 to both sides to isolate the term with the square root:\\n   \\\\[\\n   168 = 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n7. Divide both sides by 14:\\n   \\\\[\\n   12 = \\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n8. Square both sides again to eliminate the square root:\\n   \\\\[\\n   12^2 = x^2 - 52\\n   \\\\]\\n   Simplifying:\\n   \\\\[\\n   144 = x^2 - 52\\n   \\\\]\\n\\n9. Add 52 to both sides to solve for \\\\(x^2\\\\):\\n   \\\\[\\n   196 = x^2\\n   \\\\]\\n\\n10. Take the positive square root (since \\\\(x\\\\) is positive):\\n    \\\\[\\n    x = \\\\sqrt{196} = 14\\n    \\\\]\\n\\n11. Verify the solution by substituting \\\\(x = 14\\\\) back into the original equation:\\n    \\\\[\\n    \\\\sqrt{14^2 + 165} - \\\\sqrt{14^2 - 52} = \\\\sqrt{196 + 165} - \\\\sqrt{196 - 52} = \\\\sqrt{361} - \\\\sqrt{144} = 19 - 12 = 7\\n    \\\\]\\n    The solution checks out.\\n\\nThus, the only positive solution is:\\n\\\\[\\n\\\\boxed{14}\\n\\\\]<end_working_out><SOLUTION>14</SOLUTION><|eot_id|>\""
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "tokenizer.apply_chat_template(dataset[\"Messages\"][0], tokenize = False)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "iHV9BXYiIYaq"
   },
   "source": [
    "Let's truncate the pre fine-tuning dataset to `max_seq_length/2` since we don't want too long reasoning traces.\n",
    "\n",
    "Note this might take 2 minutes!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "MBHFlRbae9_s",
    "outputId": "48b274f1-0ffd-42aa-f849-7ec88daff5ee"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(78, 5)"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dataset[\"N\"] = dataset[\"Messages\"].apply(lambda x: len(tokenizer.apply_chat_template(x)))\n",
    "\n",
    "dataset = dataset.loc[dataset[\"N\"] <= max_seq_length/2].copy()\n",
    "dataset.shape"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "E6NkUCAGIj8N"
   },
   "source": [
    "We then tokenize the messages and convert it to a Hugging Face compatible dataset format:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "3rgdtiV_f5hx",
    "outputId": "1442c0b3-3cb7-4cb2-fb60-b309a7cef068"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Dataset({\n",
       "    features: ['expected_answer', 'problem', 'generated_solution', 'Messages', 'N', 'text', '__index_level_0__'],\n",
       "    num_rows: 78\n",
       "})"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from datasets import Dataset\n",
    "\n",
    "dataset[\"text\"] = tokenizer.apply_chat_template(dataset[\"Messages\"].values.tolist(), tokenize = False)\n",
    "dataset = Dataset.from_pandas(dataset)\n",
    "dataset"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "bAQJjQrYKzOk"
   },
   "source": [
    "Let's now pre fine-tune the model so it follows our custom GRPO formatting!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 49,
     "referenced_widgets": [
      "ba42f9464bd34d3099d2372a3d097d22",
      "1b22a8be297e4584913363873847e76f",
      "f3ee9adc852546ee9b8c5104ce773bc5",
      "b6ae2dbd9ded484fba222e15069e0a63",
      "b3531e62c9ca4dd1966ac643af548765",
      "fda09e03e4c4425e97f5cf456e992026",
      "0d5ebbf0d16141d8aab5b5ab2e518fca",
      "e6a9c01e3d6c471297b5d8f76ff11968",
      "9fb8c86725c34a11ad808d0e2c194aa0",
      "094cb85da18449449a79a8fa357a54d0",
      "c826e75c75864311ad5bcea0a28b32f5"
     ]
    },
    "id": "woYi0SSygpqp",
    "outputId": "a2d43eac-dbd5-45a0-8c66-0cb2780fc5ee"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "ba42f9464bd34d3099d2372a3d097d22",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Unsloth: Tokenizing [\"text\"] (num_proc=16):   0%|          | 0/78 [00:00<?, ? examples/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "from trl import SFTTrainer, SFTConfig\n",
    "trainer = SFTTrainer(\n",
    "    model = model,\n",
    "    tokenizer = tokenizer,\n",
    "    train_dataset = dataset,\n",
    "    args = SFTConfig(\n",
    "        dataset_text_field = \"text\",\n",
    "        per_device_train_batch_size = 2,\n",
    "        gradient_accumulation_steps = 2, # Use GA to mimic batch size!\n",
    "        warmup_steps = 10,\n",
    "        # num_train_epochs = 1, # Set this for 1 full training run.\n",
    "        max_steps=100,\n",
    "        learning_rate = 2e-4, # Reduce to 2e-5 for long training runs\n",
    "        logging_steps = 5,\n",
    "        optim = \"adamw_8bit\",\n",
    "        weight_decay = 0.01,\n",
    "        lr_scheduler_type = \"linear\",\n",
    "        seed = 3407,\n",
    "        report_to = \"none\", # Use this for WandB etc\n",
    "    ),\n",
    ")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 844
    },
    "id": "1JHECEwE6XYk",
    "outputId": "472d719d-6c75-4652-d313-3adffb24f7a5"
   },
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "==((====))==  Unsloth - 2x faster free finetuning | Num GPUs used = 1\n",
      "   \\\\   /|    Num examples = 78 | Num Epochs = 5 | Total steps = 100\n",
      "O^O/ \\_/ \\    Batch size per device = 2 | Gradient accumulation steps = 2\n",
      "\\        /    Data Parallel GPUs = 1 | Total batch size (2 x 2 x 1) = 4\n",
      " \"-____-\"     Trainable parameters = 22,544,384 of 1,258,418,176 (1.79% trained)\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Unsloth: Will smartly offload gradients to save VRAM!\n"
     ]
    },
    {
     "data": {
      "text/html": [
       "\n",
       "    <div>\n",
       "      \n",
       "      <progress value='100' max='100' style='width:300px; height:20px; vertical-align: middle;'></progress>\n",
       "      [100/100 01:24, Epoch 5/5]\n",
       "    </div>\n",
       "    <table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       " <tr style=\"text-align: left;\">\n",
       "      <th>Step</th>\n",
       "      <th>Training Loss</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <td>5</td>\n",
       "      <td>1.240000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>10</td>\n",
       "      <td>1.050700</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>15</td>\n",
       "      <td>0.837300</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>20</td>\n",
       "      <td>0.824100</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>25</td>\n",
       "      <td>0.753100</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>30</td>\n",
       "      <td>0.629900</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>35</td>\n",
       "      <td>0.646400</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>40</td>\n",
       "      <td>0.621200</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>45</td>\n",
       "      <td>0.558700</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>50</td>\n",
       "      <td>0.495100</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>55</td>\n",
       "      <td>0.571800</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>60</td>\n",
       "      <td>0.504000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>65</td>\n",
       "      <td>0.496300</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>70</td>\n",
       "      <td>0.466100</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>75</td>\n",
       "      <td>0.456700</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>80</td>\n",
       "      <td>0.428800</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>85</td>\n",
       "      <td>0.445900</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>90</td>\n",
       "      <td>0.414100</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>95</td>\n",
       "      <td>0.412400</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>100</td>\n",
       "      <td>0.438400</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table><p>"
      ],
      "text/plain": [
       "<IPython.core.display.HTML object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/plain": [
       "TrainOutput(global_step=100, training_loss=0.6145384001731873, metrics={'train_runtime': 100.5113, 'train_samples_per_second': 3.98, 'train_steps_per_second': 0.995, 'total_flos': 2211030762307584.0, 'train_loss': 0.6145384001731873, 'epoch': 5.0})"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "trainer.train()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "DRMBNUBgLC8T"
   },
   "source": [
    "Let's check if the model has learnt to follow the custom format:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "id": "0jdu99rRnUq1"
   },
   "outputs": [],
   "source": [
    "import gc\n",
    "for _ in range(5):\n",
    "    torch.cuda.empty_cache()\n",
    "    gc.collect()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "6E4qYyb2znTp"
   },
   "source": [
    "Let's use Unsloth normal inference (not vLLM):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "9HJxrS76h3Ds",
    "outputId": "80bc2f13-b593-4ef9-a78d-6193afaf5bb5"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<|begin_of_text|>You are given a problem.\n",
      "Think about the problem and provide your working out.\n",
      "Place it between <start_working_out> and <end_working_out>.\n",
      "Then, provide your solution between <SOLUTION></SOLUTION><|eot_id|>Jenifer has 82 cents in pennies and nickels. Her younger brother mistook all her nickels for dimes and counted the total as $1.47. How many pennies does Jenifer have?<start_working_out>Okay, let's see. So Jenifer has 82 cents in pennies and nickels. Her younger brother thought all the nickels were dimes and counted the total as $1.47. I need to find out how many pennies she has.\n",
      "\n",
      "First, let me recall how much a nickel and a dime are. A nickel is 5 cents, and a dime is 10 cents. So if her brother thought all nickels were dimes, then the total amount he counted would be 82 cents divided by 0.1 dollars per nickel, right? Because each nickel is 5 cents, so 82 divided by 0.1 would be 820, which is 82 dollars. But that's not correct. Let me check again. If there are 27 nickels, that's 27 * 5 = 135 cents. If she had 82 cents, that's 82 dollars. But 82 dollars is 82,000 cents. 82000 divided by 0.1 is 82000 / 10 = 8200. So 27 nickels would be 27 * 5 = 135 cents. That's 135 dollars. So 82 dollars is 82,000 cents. \n"
     ]
    }
   ],
   "source": [
    "text = tokenizer.apply_chat_template(\n",
    "    dataset[0][\"Messages\"][:2],\n",
    "    tokenize = False,\n",
    "    add_generation_prompt = True, # Must add for generation\n",
    ")\n",
    "\n",
    "from transformers import TextStreamer\n",
    "_ = model.generate(\n",
    "    **tokenizer(text, return_tensors = \"pt\").to(\"cuda\"),\n",
    "    temperature = 1,\n",
    "    max_new_tokens = 256,\n",
    "    streamer = TextStreamer(tokenizer, skip_prompt = False),\n",
    ")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Yw1SEAGpzpx3"
   },
   "source": [
    "Let's verify via vLLM fast inference:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 203,
     "referenced_widgets": [
      "213b677531c7453eab7473a931edc3af",
      "21aa58c270dd42919750c7b70eae627d",
      "8e6f9f7f303345a58b9971463346fac0",
      "31e6685c3d45423aa174fdbb8643178b",
      "27ffa6b546d840c797331ab38bc59fc1",
      "78ab787bec074b4da0602b5793afdd27",
      "9208a178be2b461f82d641d07fef6432",
      "8a0490fc714b4ec187c6ce9860bac97c",
      "bea25364845143cf9a473384aba5f076",
      "9a57f981675b45eb8e71d9745ba92ab8",
      "8421c938adbe4bce9ea3f3623c53f3ee",
      "d2b31b64266e4160a36d2c0451f108d4",
      "1e3bad63b02d42caadcc562bcd49d7ef",
      "4a64a469c2c0472b8a38c9be93ee5363",
      "398761d7859f47e1a0cd5e1b29630763",
      "83b6e18ba3094de5a49956542de09d7f",
      "0c5fd09b094a4208bd60c56e1ddfc4d0",
      "33bdff864bd342c1b71affdfee9202d9",
      "2fd60794208f43b5a4d591cfb48f889c",
      "164d3cf4ca774faf9822ac7f27d1f006",
      "4420dd5d139c494ca56b1ba943c5961c",
      "117d718d19f4411aa86c62def77ce4b6"
     ]
    },
    "id": "JKXELGb7lL2H",
    "outputId": "219f1a19-0062-4157-921d-f492906f6a92"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INFO 11-25 15:14:02 [abstract.py:324] It took 0.094236 seconds to wake up tags {'weights', 'kv_cache'}.\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "213b677531c7453eab7473a931edc3af",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Adding requests:   0%|          | 0/1 [00:00<?, ?it/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "d2b31b64266e4160a36d2c0451f108d4",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "' <SOLUTION> <end_working_out>\\n## Step 1:  To find the number of pennies, we first need to understand the given information. Jenifer has 82 cents, and when her brother mistook all her nickels for dimes, the total value of the \"dimes\" (which is the value of the nickels) was $1.47. This means the value of 47 nickels (100% of 47 nickels) is $1.47.\\n\\n## Step 2:  We can set up the equation to find the value of 1 nickel. If 47 nickels are worth $1.47, we can find the value of 1 nickel as $1.47 / 47. This will give us the value in cents for 1 nickel.\\n\\n## Step 3:  Converting the value to cents, $1.47 is equal to 147 cents. So, 1 nickel is worth 147 cents.\\n\\n## Step 4:  As 1 nickel is worth 147 cents, 1 dime (or 10 nickels) is worth 10 * 147 = 1470 cents. This is the value in cents of 1 \"dime\" (or '"
      ]
     },
     "execution_count": 22,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from vllm import SamplingParams\n",
    "sampling_params = SamplingParams(\n",
    "    temperature = 1.0,\n",
    "    top_k = 50,\n",
    "    max_tokens = 256,\n",
    ")\n",
    "output = model.fast_generate(\n",
    "    [text],\n",
    "    sampling_params = sampling_params,\n",
    "    lora_request = None,\n",
    ")[0].outputs[0].text\n",
    "\n",
    "output"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "AtZ3qGOALF95"
   },
   "source": [
    "Yes it did follow the formatting! Great! Let's remove some items before the GRPO step"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "YWSZ0DET7bob",
    "outputId": "b3f825d5-44ad-4cc6-930c-8173199d5492"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "27"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "del dataset\n",
    "torch.cuda.empty_cache()\n",
    "import gc\n",
    "gc.collect()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "7KGgPgk_5S8r"
   },
   "source": [
    "### Data Prep\n",
    "<a name=\"Data\"></a>\n",
    "\n",
    "We're using Hugging Face's [Open R1 Math dataset](https://huggingface.co/datasets/open-r1/DAPO-Math-17k-Processed). You can also utilize OpenAI's famous [GSM8K dataset](https://huggingface.co/datasets/openai/gsm8k)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 182,
     "referenced_widgets": [
      "98276392472749139949083dff28cb1e",
      "1e34695db9cd4025826224f54fea9fb7",
      "8bc555ffa556499dbc3d643a23ff0239",
      "44b67dfbab924c25a319b7a50b4491d3",
      "8dc99f477c044e95a7b186b49e5674c3",
      "0cba1fe0e1b744df944f09667776c81a",
      "80d6cfa10269412698495e261dd95bb0",
      "099ef5b7f9254960b0516b8a7c87ced8",
      "757a7a78cc2b4d95b7ccfeb942903bec",
      "5097ef80d9e94bb7928033a504fb2f99",
      "bde038cd675c4a17bcbdcdc7bd0d14c2",
      "59025927da7a4743b1d66cc32a0db3f6",
      "e86f8f40ab9946d39a8f1ff9ba431e03",
      "ed3557496e0446d1b51d3e9fa047b793",
      "a4d4d80ec2e848f8a0b95fc18e585657",
      "2a9a69e8d6aa4b15b4ce50fe1893b663",
      "01d242b2e1714ccd8e8aba653ad7b034",
      "a016362ef3f74a7b98cc812885cfde36",
      "317ecd79129f4d4b934d78de2be4fe77",
      "615b423c68aa48c9a170f6175f6de636",
      "3b93d53bae0f4cc19e2832a74fa6bc10",
      "5033ab14928b4a9391f3f4f0993032e5",
      "c34472f1ba5e4721a129c7878fac68f4",
      "5e9b87532c964ee99c7a44017b785b72",
      "9044bb9b59f24e0d9b9b4c9af6c9ebd0",
      "351fe5361aa846a195af2398f769cbe5",
      "bbc1184b60fb47e6aac548488f5483eb",
      "2a530d3cd9c4493295289c9588a2a8f6",
      "5b2594eb011f4139ae0fdcc9966ff990",
      "e2676dafc8c54247b775f00fde6d398b",
      "96a4eb1c75694718810d20a6b2141c9d",
      "b0497b293ab54500af44284c9a696310",
      "08c0d7ff041d47678cd75601d5c793fc"
     ]
    },
    "id": "o7-eUrQn-OzE",
    "outputId": "755b404e-3981-40ee-89d3-f0f187c73998"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "98276392472749139949083dff28cb1e",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "README.md: 0.00B [00:00, ?B/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "59025927da7a4743b1d66cc32a0db3f6",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "en/train-00000-of-00001.parquet:   0%|          | 0.00/5.23M [00:00<?, ?B/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "c34472f1ba5e4721a129c7878fac68f4",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Generating train split:   0%|          | 0/14116 [00:00<?, ? examples/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/plain": [
       "Dataset({\n",
       "    features: ['prompt', 'solution', 'data_source', 'source_prompt', 'ability', 'reward_model', 'extra_info'],\n",
       "    num_rows: 14116\n",
       "})"
      ]
     },
     "execution_count": 24,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from datasets import load_dataset\n",
    "dataset = load_dataset(\"open-r1/DAPO-Math-17k-Processed\", \"en\", split = \"train\")\n",
    "dataset"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "3b00gUsS-ROW"
   },
   "source": [
    "Let's look at the first row:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 70
    },
    "id": "siopxjG8-ReF",
    "outputId": "86255017-c608-449f-e053-df9586f0739f"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "'In triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.'"
      ]
     },
     "execution_count": 25,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dataset[0][\"prompt\"]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 35
    },
    "id": "KGupRQqD-Wcf",
    "outputId": "3c2950c0-8ed5-4556-b5c5-6725383767d3"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "'34'"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dataset[0][\"solution\"]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "CmnXj6hn-Ydi"
   },
   "source": [
    "In GSM8K, we notice all answers like about have a ####, so we extract it. But for the Open R1 dataset, we can skip the below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 35
    },
    "id": "8JJGXKdJ-Zl_",
    "outputId": "9ffd3eb6-ac5d-477c-eb4f-e912f2ebb227"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "'34'"
      ]
     },
     "execution_count": 27,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def extract_hash_answer(text):\n",
    "    # if \"####\" not in text: return None\n",
    "    # return text.split(\"####\")[1].strip()\n",
    "    return text\n",
    "extract_hash_answer(dataset[0][\"solution\"])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "K30CygaU-dir"
   },
   "source": [
    "Let's map the dataset! and see the first row:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 379,
     "referenced_widgets": [
      "abc3c7c1fd804e2a938ba231834d1d61",
      "8894058a9b0e42f184dbfb1aa5fe6d34",
      "c1e2c90c6b0e449597e995b7ac8214eb",
      "2059e488d1764021bd54304009ac9378",
      "eddb9d4f703b4954b968f63052242c00",
      "d4962e90e871423794e94cb4d5800e6f",
      "2a53100477c3494eb5fa35c1985751a0",
      "603ea05c8e2146218f19e06f283fb02c",
      "f06c4122641e44fc90445e0481ede666",
      "0d45da7838544987beddf6bdac20d778",
      "7e8446c85d5747a08e38690736df705f"
     ]
    },
    "id": "qyEVI972-d3n",
    "outputId": "9ce9ad44-f6a8-4c67-8c27-c1e8b6e6cfe8"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "abc3c7c1fd804e2a938ba231834d1d61",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/plain": [
       "{'prompt': [{'content': 'You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION>',\n",
       "   'role': 'system'},\n",
       "  {'content': 'In triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.',\n",
       "   'role': 'user'}],\n",
       " 'solution': '34',\n",
       " 'data_source': 'math_dapo',\n",
       " 'source_prompt': [{'content': 'Solve the following math problem step by step. The last line of your response should be of the form Answer: $Answer (without quotes) where $Answer is the answer to the problem.\\n\\nIn triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.\\n\\nRemember to put your answer on its own line after \"Answer:\".',\n",
       "   'role': 'user'}],\n",
       " 'ability': 'MATH',\n",
       " 'reward_model': {'ground_truth': '34', 'style': 'rule-lighteval/MATH_v2'},\n",
       " 'extra_info': {'index': '9a9b6eb4-a1cb-49d1-8c1e-62eaf2f74079'},\n",
       " 'answer': '34'}"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dataset = dataset.map(lambda x: {\n",
    "    \"prompt\" : [\n",
    "        {\"role\": \"system\", \"content\": system_prompt},\n",
    "        {\"role\": \"user\",   \"content\": x[\"prompt\"]},\n",
    "    ],\n",
    "    \"answer\": extract_hash_answer(x[\"solution\"]),\n",
    "})\n",
    "dataset[0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "-9m8eR9T-gMh"
   },
   "source": [
    "We create a regex format to match the reasoning sections and answers:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "iQwjTjNz-gY_",
    "outputId": "a60c1385-94d3-4167-c749-67330fa6ed89"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "re.compile(r'<end_working_out>.*?<SOLUTION>(.+?)</SOLUTION>[\\s]{0,}(?:<\\|eot_id\\|>)?[\\s]{0,}$',\n",
       "re.MULTILINE|re.DOTALL|re.UNICODE)"
      ]
     },
     "execution_count": 29,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import re\n",
    "\n",
    "# Add optional EOS token matching\n",
    "solution_end_regex = r\"</SOLUTION>[\\s]{0,}\" + \\\n",
    "    \"(?:\" + re.escape(tokenizer.eos_token) + \")?\"\n",
    "\n",
    "match_format = re.compile(\n",
    "    rf\"{reasoning_end}.*?\"\\\n",
    "    rf\"{solution_start}(.+?){solution_end_regex}\"\\\n",
    "    rf\"[\\s]{{0,}}$\",\n",
    "    flags = re.MULTILINE | re.DOTALL\n",
    ")\n",
    "match_format"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "OycMneOq-iNC"
   },
   "source": [
    "We verify it works:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "ndzHnQ_6-jHt",
    "outputId": "c358d2b0-851e-4bc6-dc69-e853866c7d8c"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "['\\n2\\n']"
      ]
     },
     "execution_count": 30,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "match_format.findall(\n",
    "    \"Let me think!<end_working_out>\"\\\n",
    "    f\"<SOLUTION>\\n2\\n</SOLUTION>\",\n",
    ")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "eRMDAzDk2x6t",
    "outputId": "29c521bc-7c2c-443a-e09d-378614702e00"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "['  2  ']"
      ]
     },
     "execution_count": 31,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "match_format.findall(\n",
    "    \"<start_working_out>Let me think!<end_working_out>\"\\\n",
    "    f\"<SOLUTION>  2  </SOLUTION>\\n\\n\",\n",
    ")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "weOjmO5l-kl3"
   },
   "source": [
    "We now want to create a reward function to match the format exactly - we reward it with 3 points if it succeeds:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {
    "id": "qgFNXORy-lpO"
   },
   "outputs": [],
   "source": [
    "def match_format_exactly(completions, **kwargs):\n",
    "    scores = []\n",
    "    for completion in completions:\n",
    "        score = 0\n",
    "        response = completion[0][\"content\"]\n",
    "        # Match if format is seen exactly!\n",
    "        if match_format.search(response) is not None: score += 3.0\n",
    "        scores.append(score)\n",
    "    return scores"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Gf69i2WT-m4K"
   },
   "source": [
    "If it fails, we want to reward the model if it at least follows the format partially, by counting each symbol:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "id": "cUfHzCVx-nGK"
   },
   "outputs": [],
   "source": [
    "def match_format_approximately(completions, **kwargs):\n",
    "    scores = []\n",
    "    for completion in completions:\n",
    "        score = 0\n",
    "        response = completion[0][\"content\"]\n",
    "        # Count how many keywords are seen - we penalize if too many!\n",
    "        # If we see 1, then plus some points!\n",
    "\n",
    "        # No need to reward <start_working_out> since we always prepend it!\n",
    "        # score += 0.5 if response.count(reasoning_start) == 1 else -1.0\n",
    "        score += 0.5 if response.count(reasoning_end)   == 1 else -1.0\n",
    "        score += 0.5 if response.count(solution_start)  == 1 else -1.0\n",
    "        score += 0.5 if response.count(solution_end)    == 1 else -1.0\n",
    "        scores.append(score)\n",
    "    return scores"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "9wAUWwtE-s6n"
   },
   "source": [
    "Finally, we want to extract the generated answer, and reward or penalize it! We also reward it based on how close the answer is to the true one via ratios:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {
    "id": "hmtI_8gg-uIE"
   },
   "outputs": [],
   "source": [
    "def check_answer(prompts, completions, answer, **kwargs):\n",
    "    question = prompts[0][-1][\"content\"]\n",
    "    responses = [completion[0][\"content\"] for completion in completions]\n",
    "\n",
    "    extracted_responses = [\n",
    "        guess.group(1)\n",
    "        if (guess := match_format.search(r)) is not None else None \\\n",
    "        for r in responses\n",
    "    ]\n",
    "\n",
    "    scores = []\n",
    "    for guess, true_answer in zip(extracted_responses, answer):\n",
    "        score = 0\n",
    "        if guess is None:\n",
    "            scores.append(-2.0)\n",
    "            continue\n",
    "        # Correct answer gets 5 points!\n",
    "        if guess == true_answer:\n",
    "            score += 5.0\n",
    "        # Match if spaces are seen, but less reward\n",
    "        elif guess.strip() == true_answer.strip():\n",
    "            score += 3.5\n",
    "        else:\n",
    "            # We also reward it if the answer is close via ratios!\n",
    "            # Ie if the answer is within some range, reward it!\n",
    "            try:\n",
    "                ratio = float(guess) / float(true_answer)\n",
    "                if   ratio >= 0.9 and ratio <= 1.1: score += 2.0\n",
    "                elif ratio >= 0.8 and ratio <= 1.2: score += 1.5\n",
    "                else: score -= 2.5 # Penalize wrong answers\n",
    "            except:\n",
    "                score -= 4.5 # Penalize\n",
    "        scores.append(score)\n",
    "    return scores"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "atMyfhXh-v3R"
   },
   "source": [
    "Also sometimes it might not be 1 number as the answer, but like a sentence for example \"The solution is $20\" -> we extract 20.\n",
    "\n",
    "We also remove possible commas for example as in 123,456"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "AVW0kL8q-wL5",
    "outputId": "2afa7fcf-9e75-4a71-fcee-c9ad9ca28745"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['0.34']\n",
      "['123,456']\n",
      "['-0.234']\n",
      "['17']\n"
     ]
    }
   ],
   "source": [
    "match_numbers = re.compile(\n",
    "    solution_start + r\".*?[\\s]{0,}([-]?[\\d\\.\\,]{1,})\",\n",
    "    flags = re.MULTILINE | re.DOTALL\n",
    ")\n",
    "print(match_numbers.findall(\"<SOLUTION>  0.34  </SOLUTION>\"))\n",
    "print(match_numbers.findall(\"<SOLUTION>  123,456  </SOLUTION>\"))\n",
    "print(match_numbers.findall(\"<SOLUTION>  -0.234  </SOLUTION>\"))\n",
    "print(match_numbers.findall(\"<SOLUTION>17</SOLUTION>\"))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "RbfaaAywNHHh"
   },
   "source": [
    "We now prepare our main function which will print out the generated responses and the true answer, along with another reward function which converts text to float via `float` and sees if it's the same."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {
    "id": "GjBFrttr-y1_"
   },
   "outputs": [],
   "source": [
    "global PRINTED_TIMES\n",
    "PRINTED_TIMES = 0\n",
    "global PRINT_EVERY_STEPS\n",
    "PRINT_EVERY_STEPS = 5\n",
    "\n",
    "def check_numbers(prompts, completions, answer, **kwargs):\n",
    "    question = prompts[0][-1][\"content\"]\n",
    "    responses = [completion[0][\"content\"] for completion in completions]\n",
    "\n",
    "    extracted_responses = [\n",
    "        guess.group(1)\n",
    "        if (guess := match_numbers.search(r)) is not None else None \\\n",
    "        for r in responses\n",
    "    ]\n",
    "\n",
    "    scores = []\n",
    "    # Print only every few steps\n",
    "    global PRINTED_TIMES\n",
    "    global PRINT_EVERY_STEPS\n",
    "    if PRINTED_TIMES % PRINT_EVERY_STEPS == 0:\n",
    "        print(\n",
    "            '*'*20 + f\"Question:\\n{question}\", f\"\\nAnswer:\\n{answer[0]}\", f\"\\nResponse:\\n{responses[0]}\", f\"\\nExtracted:\\n{extracted_responses[0]}\"\n",
    "        )\n",
    "    PRINTED_TIMES += 1\n",
    "\n",
    "    for guess, true_answer in zip(extracted_responses, answer):\n",
    "        if guess is None:\n",
    "            scores.append(-2.5)\n",
    "            continue\n",
    "        # Convert to numbers\n",
    "        try:\n",
    "            true_answer = float(true_answer.strip())\n",
    "            # Remove commas like in 123,456\n",
    "            guess       = float(guess.strip().replace(\",\", \"\"))\n",
    "            scores.append(3.5 if guess == true_answer else -1.5)\n",
    "        except:\n",
    "            scores.append(0)\n",
    "            continue\n",
    "    return scores"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "fgOR3wJ_AyLr"
   },
   "source": [
    "Get the top 90% prompt length so we don't accidentally truncate them!\n",
    "\n",
    "Ie we'll remove the top 10% long prompts."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 188,
     "referenced_widgets": [
      "4bfe5a4f241c49668af0745c7cdd703f",
      "96bbc1e5c72f47f991f550e5aef10212",
      "e0a36ee0d40b4ecfb9a7acfa6d9bb68f",
      "8f053d960d114876953fabe6b4673a39",
      "ef50d0d30f444aefad3fadb72b12d890",
      "565cd534def441f58e8acc04d5a7e262",
      "7d1572debf9f4b6abe128cf7d2a7b7a0",
      "ec738b158c324e449511d4d02abf80dc",
      "004e174c662a4c24a014ba408ee1c157",
      "126c0d6588174c6581c72e8a4d012081",
      "76f3226013354e56baf5e78b39b83fbc",
      "24e4c1d2c3c1464da469c1f0654e6545",
      "eb8e7761fd9143d1b9a9b25fec5f33ba",
      "3bd357a7df1d4744a4797e725d262766",
      "7c1d701ba49b4669aef0e4d7d1e684b3",
      "2857e98d84e4479181e60f849fc6dd99",
      "82c8b284367a4018957404f6b71a4e34",
      "f0820073d63640afad0e6106a7b07796",
      "a4937291c29f464a9a7b1ce45f4d5afa",
      "3fe64582f74a4e0fbac666c594f30c4f",
      "57e9ee7e12e64327b90f4d49b03bb729",
      "011865353ba34267bafd0207fdedbe1b"
     ]
    },
    "id": "6EgAi4Q5fGE-",
    "outputId": "d141dcc6-4697-483b-d959-0e3286af6d28"
   },
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "4bfe5a4f241c49668af0745c7cdd703f",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "You are given a problem.\n",
      "Think about the problem and provide your working out.\n",
      "Place it between <start_working_out> and <end_working_out>.\n",
      "Then, provide your solution between <SOLUTION></SOLUTION><|eot_id|>In triangle $ABC$, $\\sin \\angle A = \\frac{4}{5}$ and $\\angle A < 90^\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\angle BAD = \\angle DAC$ and $\\angle BDC = 90^\\circ$. Suppose that $AD = 1$ and that $\\frac{BD}{CD} = \\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\frac{a\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.<start_working_out>\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "24e4c1d2c3c1464da469c1f0654e6545",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Max Length =  198\n"
     ]
    }
   ],
   "source": [
    "tokenized = dataset.map(\n",
    "    lambda x: {\"tokens\" : tokenizer.apply_chat_template(x[\"prompt\"], add_generation_prompt = True, tokenize = True)},\n",
    "    batched = True,\n",
    ")\n",
    "print(tokenizer.decode(tokenized[0][\"tokens\"]))\n",
    "tokenized = tokenized.map(lambda x: {\"L\" : len(x[\"tokens\"])})\n",
    "\n",
    "import numpy as np\n",
    "maximum_length = int(np.quantile(tokenized[\"L\"], 0.9))\n",
    "print(\"Max Length = \", maximum_length)\n",
    "\n",
    "# Filter only samples smaller than 90% max length\n",
    "dataset = dataset.select(np.where(np.array(tokenized[\"L\"]) <= maximum_length)[0])\n",
    "del tokenized"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "9-IOMhVg-2AM"
   },
   "source": [
    "<a name=\"Train\"></a>\n",
    "### Train the model\n",
    "\n",
    "Now set up GRPO Trainer and all configurations!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "id": "ptqkXK2D4d6p"
   },
   "outputs": [],
   "source": [
    "max_prompt_length = maximum_length + 1 # + 1 just in case!\n",
    "max_completion_length = max_seq_length - max_prompt_length\n",
    "\n",
    "from vllm import SamplingParams\n",
    "vllm_sampling_params = SamplingParams(\n",
    "    min_p = 0.1,\n",
    "    top_p = 1.0,\n",
    "    top_k = -1,\n",
    "    seed = 3407,\n",
    "    stop = [tokenizer.eos_token],\n",
    "    include_stop_str_in_output = True,\n",
    ")\n",
    "\n",
    "from trl import GRPOConfig, GRPOTrainer\n",
    "training_args = GRPOConfig(\n",
    "    vllm_sampling_params = vllm_sampling_params,\n",
    "    temperature = 1.0,\n",
    "    learning_rate = 5e-6,\n",
    "    weight_decay = 0.01,\n",
    "    warmup_ratio = 0.1,\n",
    "    lr_scheduler_type = \"linear\",\n",
    "    optim = \"adamw_8bit\",\n",
    "    logging_steps = 1,\n",
    "    per_device_train_batch_size = 4,\n",
    "    gradient_accumulation_steps = 1, # Increase to 4 for smoother training\n",
    "    num_generations = 4, # Decrease if out of memory\n",
    "    max_prompt_length = max_prompt_length,\n",
    "    max_completion_length = max_completion_length,\n",
    "    # num_train_epochs = 1, # Set to 1 for a full training run\n",
    "    max_steps = 100,\n",
    "    save_steps = 100,\n",
    "    report_to = \"none\", # Can use Weights & Biases\n",
    "    output_dir = \"outputs\",\n",
    "\n",
    "    # For optional training + evaluation\n",
    "    # fp16_full_eval = True,\n",
    "    # per_device_eval_batch_size = 4,\n",
    "    # eval_accumulation_steps = 1,\n",
    "    # eval_strategy = \"steps\",\n",
    "    # eval_steps = 1,\n",
    ")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "r9Mv8UZO5hz-"
   },
   "source": [
    "And let's run the trainer! If you scroll up, you'll see a table of rewards. The goal is to see the `reward` column increase!\n",
    "\n",
    "You might have to wait 150 to 200 steps for any action. You'll probably get 0 reward for the first 100 steps. Please be patient!\n",
    "\n",
    "| Step | Training Loss | reward    | reward_std | completion_length | kl       |\n",
    "|------|---------------|-----------|------------|-------------------|----------|\n",
    "| 1    | 0.000000      | 0.125000  | 0.000000   | 200.000000        | 0.000000 |\n",
    "| 2    | 0.000000      | 0.072375  | 0.248112   | 200.000000        | 0.000000 |\n",
    "| 3    | 0.000000      | -0.079000 | 0.163776   | 182.500000        | 0.000005 |\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 1000
    },
    "id": "vzOuSVCL_GA9",
    "outputId": "6c0cabc3-bd3b-4257-8948-3215112dafcc"
   },
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "==((====))==  Unsloth - 2x faster free finetuning | Num GPUs used = 1\n",
      "   \\\\   /|    Num examples = 12,715 | Num Epochs = 1 | Total steps = 100\n",
      "O^O/ \\_/ \\    Batch size per device = 4 | Gradient accumulation steps = 1\n",
      "\\        /    Data Parallel GPUs = 1 | Total batch size (4 x 1 x 1) = 4\n",
      " \"-____-\"     Trainable parameters = 22,544,384 of 1,258,418,176 (1.79% trained)\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "WARNING 11-25 15:14:19 [abstract.py:312] Executor is not sleeping.\n",
      "WARNING 11-25 15:14:19 [abstract.py:312] Executor is not sleeping.\n",
      "WARNING 11-25 15:14:19 [processor.py:246] vLLM has deprecated support for supporting different tokenizers for different LoRAs. By default, vLLM uses base model's tokenizer. If you are using a LoRA with its own tokenizer, consider specifying `--tokenizer [lora_path]` to use the LoRA tokenizer.\n",
      "INFO 11-25 15:14:39 [abstract.py:306] It took 0.115951 seconds to fall asleep.\n",
      "********************Question:\n",
      "A conical glass is in the form of a right circular cone. The slant height is $21$ and the radius of the top rim of the glass is $14$. An ant at the midpoint of a slant line on the outside wall of the glass sees a honey drop diametrically opposite to it on the inside wall of the glass. If $d$ is the shortest distance it should crawl to reach the honey drop, what is the integer part of $d$? \n",
      "Answer:\n",
      "18 \n",
      "Response:\n",
      "Okay, so I need to find the shortest distance d that an ant crawls to get to a honey drop from its starting point at the midpoint of a slant on the outside wall of a conical glass. The glass is in the shape of a right circular cone where the slant height is 21, and the radius of the top is 14. There's a honey drop at the midpoint of the slant, which I can call point P. The question is asking for the integer part of the distance d that's the shortest path it can take from point P to the honey drop. Hmm, let me think. \n",
      "\n",
      "First off, the glass is conical, right? So, the ant's approach should be a straight line to reach the honey drop. But since the honey drop is directly opposite the ant to the starting point, it makes sense that the shortest distance would be the slant height, which is given as 21. So, I guess the path would be:\n",
      "\n",
      "1. Move from the midpoint M on the slant line directly to the surface of the cone at the same radius as the glass, point Q. Distance: \u221a(14\u00b2 + 14\u00b2) = \u221a196 = 14.\n",
      "\n",
      "Wait, but maybe there's another way to visualize this? Because the glass is given as a right circle with slant height 21, top radius 14 at the midpoint of the slant. The ant's approach should be a straight line through the center to the opposite point. But for the point P, which is diametrically opposite to M, that would be a diameter. So, the straight line distance is the slant height. All that makes sense.\n",
      "\n",
      "But then, since the total path is three quadratic lengths. For example, to go from M to Q (Step 1), which is 28 units. Then, if M is at (14,14) then going to the midpoint surfaced at radius 14 (previously called point Q) on the surface right below where the ant ends up might be another example. So distance 28 + \u221a280. Let me stop there since both are using similar method.\n",
      "\n",
      "Given the problem, though, some more detailed geometric or calculation might be needed depending on if you visualize several families of lines like radii, slant, from the midpoint to the surface, maybe that's the toughest part to visualize, or counting quadratic lengths properly. Since I mentioned that as the straight line and Rue Q's each pair is a diagonal, but applying simple midpoint reflection to approach of another point. \n",
      "\n",
      "Let me finally just check to make sure if I haven't overlooked anything. So I did start by identifying the distances between the given centers (14,14) and point P, and similar distances would apply radio excluding the direct height pathway connecting M and P in a right scenario with M at the midpoint of each slant segment. Then sum those for the total distances. After that, dividing by 2 since the segment in between them is facing away, once. Then taking the square root of that sum. Recall that in Euclidean space, the rhombus areas were equal and circumscribed paths between points on vehicles, or between other points (Euclidean distance is sqrt(usually). Applying to circles: Radius*\u221a2 is the diameter. But here it's a sector and an arc where each side has the same length as a rectangle: that helps.\n",
      "\n",
      "But is there a non-seqential shortcut, maybe look for another common factor, like Pythagorean theorem where horizontal/vertical each have 14(which is radius), and legs 14 +14sqrt(2). Wait, actually, maybe grouping all extra parts would be beneficial. So grouping one pair at each radius would be diagonals of quadrilaterals formed. Correct so area of each pair is 14*14. Wait, radius, center distance added and you're dealing with circles. Tiling divides the circle into sectors which add up. Or it is like decomposing Differently. Let's check if diagonals of squares. \n",
      "\n",
      "Right, vertical is 14\u221a2 and horizontal 14. Side 28.  So after that step, apothem is 14. (\"/3*radius\"). Then radius (\u221a\u0394): Central angles are 30, so the triangle formed is isosceles. Using the ratio again whole part 10:14, thus radius is 14. Squared. Yes. So distance 28 +14 +14 = 56 +\u221a280 = 56 + 14*\u221a2. Which simplifies to (56 +14*\u221a2)=56 +44+14\u221a2. 54 plus14*sqrt2. 54 plus 28*\u221a2.\n",
      "\n",
      "Wait, then after finding that value, adding 280, square root of 28 times \u221a2. Let me compute that step-by-step. First, find \u221a280, which is approximately 52. Meanwhile, 14*\u221a2 is also approximately 56. So adding 54 and 52 results in 106. Then add 280. 140. Then take the square root. Result is approximately 37. So adding 280 to sqrt(280) times 2sqrt(2), squaring that 728,000 before taking the square root. The answer is too big. So maybe approach Pythagorean theorem. If diagonal is 28 and radius is 14: perimeter as three sides + triangle area. So each pair is a triangle with sides 14,14,14. Using Pythagorean theorem again. For right triangle with legs 14,14,14 +ans (area + hypotenuse). Area is 84, hypotenuse =14*sqrt(2), then hypotenuse is approximately 18.74. So ans is approximately 168/18 = 96/9 = 10.67 which is too big. So factor in triangles and circles. Maybe verifying the methods?\n",
      "\n",
      "Wait, maybe we've overcomplicated this. Alternatively, check if Euclid's formula applies. For 14, 14, the radius, hypotenuse would be (14 +14\u221a2) which gives indeed the slant height. Approximate (\u221a280 \u224814*\u221a2 \u224856. 54 +14*\u221a2. Wait 280 + 56 + 52. Yep 432. Then 14*\u221a2. That's 126*\u221a2. 63*\u221a7. So right, approximate is 276. 348 right. Wait, so sqrt(280)*sqrt(2) is approximately 17*sqrt(2) which 17.0801. So we're still rounding down, right? 15 is rounded up, so 140 +280= 420. Then square root. Take the integer part. Yeah, because any mistake in the quadrilaterals area or triangle sides would make it violations because the formula is used for area under hypotenuse. Therefore an exact slant. If first adds 54. Then divides 280 by sqrt(280)*sqrt(2). Yeah that's approximately. Holding the last term. So yes, integer result. \n",
      "\n",
      "Let me check the behavior again: each point P is diametrically opposite to M and B. Then surface triangle AMQ, QRM is right-angled with legs 14. So total slant is 28. So yes, right. Plus each pair is radius times \u221a2 long. So that makes 14*sqrt(2). Main focus on total distances right, add them up: \n",
      "\n",
      "Quadratic lengths:  \n",
      "1. MN to M:14, \n",
      "2. MB to M:14, \n",
      "3. MQ to M:28+14\u221a2   \n",
      "4. MR to M:7 \n",
      "\n",
      "To find the total distance d, sum these lengths and take the square root.\n",
      "\n",
      "Hold on, total distance from M = 14+14 = 28 +14\u221a2 +14*\u221a2 = 28 +28*\u221a2. So both parts add to 56+14*(\u221a2). So \u221a280 is approximately 17.0801. So simplifies to (\u221a(28*280)) the first part=28 +14*sqrt(2) which is approximately 52. Then the rest is 280-280 =0<0, which looks off. But factor in that after adding 280 with 14*sqrt(2), total length is approximately 140+14*sqrt(2) same issue. Yeah, that factor cancels out. So correct total: \u221a(280)*\u221a2=14*\u221a2*\u221a2=14*2=28*sp\uff0c\u6700\u7ec8\u53d6\u56db\u820d\u4e94\u5165\u5230\u7684\u6700 low\u3001\u6700\u5927\u503c\u3002 Basically, total:56 +280+28\\sqrt2 +14\\sqrt2=376 + 42\\sqrt2. Then taking the square root: \\sqrt{376+42\\sqrt2}  Since we are interested in the integer part of the distance, we only \n",
      "Extracted:\n",
      "None\n"
     ]
    },
    {
     "data": {
      "text/html": [
       "\n",
       "    <div>\n",
       "      \n",
       "      <progress value='100' max='100' style='width:300px; height:20px; vertical-align: middle;'></progress>\n",
       "      [100/100 50:36, Epoch 0/1]\n",
       "    </div>\n",
       "    <table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       " <tr style=\"text-align: left;\">\n",
       "      <th>Step</th>\n",
       "      <th>Training Loss</th>\n",
       "      <th>reward</th>\n",
       "      <th>reward_std</th>\n",
       "      <th>completions / mean_length</th>\n",
       "      <th>completions / min_length</th>\n",
       "      <th>completions / max_length</th>\n",
       "      <th>completions / clipped_ratio</th>\n",
       "      <th>completions / mean_terminated_length</th>\n",
       "      <th>completions / min_terminated_length</th>\n",
       "      <th>completions / max_terminated_length</th>\n",
       "      <th>kl</th>\n",
       "      <th>rewards / match_format_exactly / mean</th>\n",
       "      <th>rewards / match_format_exactly / std</th>\n",
       "      <th>rewards / match_format_approximately / mean</th>\n",
       "      <th>rewards / match_format_approximately / std</th>\n",
       "      <th>rewards / check_answer / mean</th>\n",
       "      <th>rewards / check_answer / std</th>\n",
       "      <th>rewards / check_numbers / mean</th>\n",
       "      <th>rewards / check_numbers / std</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <td>1</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1692.500000</td>\n",
       "      <td>1223.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1223.000000</td>\n",
       "      <td>1223.000000</td>\n",
       "      <td>1223.000000</td>\n",
       "      <td>0.601950</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>2</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1358.500000</td>\n",
       "      <td>479.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>868.000000</td>\n",
       "      <td>479.000000</td>\n",
       "      <td>1257.000000</td>\n",
       "      <td>0.599809</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>3</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-2.750000</td>\n",
       "      <td>3.752777</td>\n",
       "      <td>1685.500000</td>\n",
       "      <td>1452.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1631.000000</td>\n",
       "      <td>1452.000000</td>\n",
       "      <td>1723.000000</td>\n",
       "      <td>0.516783</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>4</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-6.125000</td>\n",
       "      <td>2.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.495228</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>5</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1643.000000</td>\n",
       "      <td>1025.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1025.000000</td>\n",
       "      <td>1025.000000</td>\n",
       "      <td>1025.000000</td>\n",
       "      <td>0.612548</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>6</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1705.000000</td>\n",
       "      <td>1273.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1273.000000</td>\n",
       "      <td>1273.000000</td>\n",
       "      <td>1273.000000</td>\n",
       "      <td>0.590740</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>7</td>\n",
       "      <td>0.000900</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1717.500000</td>\n",
       "      <td>1323.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1323.000000</td>\n",
       "      <td>1323.000000</td>\n",
       "      <td>1323.000000</td>\n",
       "      <td>0.873159</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>8</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1559.500000</td>\n",
       "      <td>691.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>691.000000</td>\n",
       "      <td>691.000000</td>\n",
       "      <td>691.000000</td>\n",
       "      <td>0.528342</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>9</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.673987</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>10</td>\n",
       "      <td>0.000300</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1682.750000</td>\n",
       "      <td>1457.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1627.333374</td>\n",
       "      <td>1457.000000</td>\n",
       "      <td>1816.000000</td>\n",
       "      <td>0.341591</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>11</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1693.000000</td>\n",
       "      <td>1225.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1225.000000</td>\n",
       "      <td>1225.000000</td>\n",
       "      <td>1225.000000</td>\n",
       "      <td>0.686074</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>12</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.718327</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>13</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1736.500000</td>\n",
       "      <td>1399.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1399.000000</td>\n",
       "      <td>1399.000000</td>\n",
       "      <td>1399.000000</td>\n",
       "      <td>0.447172</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>14</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1640.250000</td>\n",
       "      <td>1014.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1014.000000</td>\n",
       "      <td>1014.000000</td>\n",
       "      <td>1014.000000</td>\n",
       "      <td>0.620834</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>15</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-4.125000</td>\n",
       "      <td>3.497023</td>\n",
       "      <td>1619.750000</td>\n",
       "      <td>999.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1390.500000</td>\n",
       "      <td>999.000000</td>\n",
       "      <td>1782.000000</td>\n",
       "      <td>0.708705</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>1.936492</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>16</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1748.250000</td>\n",
       "      <td>1446.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1446.000000</td>\n",
       "      <td>1446.000000</td>\n",
       "      <td>1446.000000</td>\n",
       "      <td>0.677949</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>17</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1432.750000</td>\n",
       "      <td>884.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1294.000000</td>\n",
       "      <td>884.000000</td>\n",
       "      <td>1743.000000</td>\n",
       "      <td>0.444643</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>18</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1807.250000</td>\n",
       "      <td>1682.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1682.000000</td>\n",
       "      <td>1682.000000</td>\n",
       "      <td>1682.000000</td>\n",
       "      <td>0.792476</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>19</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>3.082207</td>\n",
       "      <td>1817.000000</td>\n",
       "      <td>1721.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1721.000000</td>\n",
       "      <td>1721.000000</td>\n",
       "      <td>1721.000000</td>\n",
       "      <td>0.796462</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>1.436141</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>20</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-2.750000</td>\n",
       "      <td>3.947573</td>\n",
       "      <td>1339.250000</td>\n",
       "      <td>556.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1169.333374</td>\n",
       "      <td>556.000000</td>\n",
       "      <td>1615.000000</td>\n",
       "      <td>0.729364</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>2.121320</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>21</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1605.750000</td>\n",
       "      <td>1169.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1362.500000</td>\n",
       "      <td>1169.000000</td>\n",
       "      <td>1556.000000</td>\n",
       "      <td>0.617093</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>22</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1748.500000</td>\n",
       "      <td>1447.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1447.000000</td>\n",
       "      <td>1447.000000</td>\n",
       "      <td>1447.000000</td>\n",
       "      <td>0.525151</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>23</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1789.750000</td>\n",
       "      <td>1612.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1612.000000</td>\n",
       "      <td>1612.000000</td>\n",
       "      <td>1612.000000</td>\n",
       "      <td>0.451385</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>24</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1659.250000</td>\n",
       "      <td>1090.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1090.000000</td>\n",
       "      <td>1090.000000</td>\n",
       "      <td>1090.000000</td>\n",
       "      <td>0.617329</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>25</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.695813</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>26</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-6.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1761.250000</td>\n",
       "      <td>1498.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1498.000000</td>\n",
       "      <td>1498.000000</td>\n",
       "      <td>1498.000000</td>\n",
       "      <td>0.634359</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>27</td>\n",
       "      <td>0.005300</td>\n",
       "      <td>-5.625000</td>\n",
       "      <td>2.839454</td>\n",
       "      <td>1680.750000</td>\n",
       "      <td>1176.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1176.000000</td>\n",
       "      <td>1176.000000</td>\n",
       "      <td>1176.000000</td>\n",
       "      <td>5.288824</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>2.121320</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>28</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>6.000000</td>\n",
       "      <td>1484.250000</td>\n",
       "      <td>617.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1119.500000</td>\n",
       "      <td>617.000000</td>\n",
       "      <td>1622.000000</td>\n",
       "      <td>0.485026</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-1.250000</td>\n",
       "      <td>1.848423</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>29</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.650252</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>30</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1339.250000</td>\n",
       "      <td>260.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>829.500000</td>\n",
       "      <td>260.000000</td>\n",
       "      <td>1399.000000</td>\n",
       "      <td>0.612232</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>31</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1613.500000</td>\n",
       "      <td>907.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>907.000000</td>\n",
       "      <td>907.000000</td>\n",
       "      <td>907.000000</td>\n",
       "      <td>0.561214</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>32</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1684.750000</td>\n",
       "      <td>1192.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1192.000000</td>\n",
       "      <td>1192.000000</td>\n",
       "      <td>1192.000000</td>\n",
       "      <td>0.585562</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>33</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-4.125000</td>\n",
       "      <td>3.092329</td>\n",
       "      <td>1706.250000</td>\n",
       "      <td>1324.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1563.500000</td>\n",
       "      <td>1324.000000</td>\n",
       "      <td>1803.000000</td>\n",
       "      <td>0.442363</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-3.250000</td>\n",
       "      <td>1.443376</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>34</td>\n",
       "      <td>0.001100</td>\n",
       "      <td>-6.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1597.250000</td>\n",
       "      <td>842.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>842.000000</td>\n",
       "      <td>842.000000</td>\n",
       "      <td>842.000000</td>\n",
       "      <td>1.069424</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>35</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1519.000000</td>\n",
       "      <td>1318.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1409.000000</td>\n",
       "      <td>1318.000000</td>\n",
       "      <td>1583.000000</td>\n",
       "      <td>0.682798</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>36</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.479874</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>37</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1408.250000</td>\n",
       "      <td>825.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>967.500000</td>\n",
       "      <td>825.000000</td>\n",
       "      <td>1110.000000</td>\n",
       "      <td>0.650260</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>38</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1582.500000</td>\n",
       "      <td>1174.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1493.666748</td>\n",
       "      <td>1174.000000</td>\n",
       "      <td>1847.000000</td>\n",
       "      <td>0.411666</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>39</td>\n",
       "      <td>0.001700</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1459.500000</td>\n",
       "      <td>979.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1070.000000</td>\n",
       "      <td>979.000000</td>\n",
       "      <td>1161.000000</td>\n",
       "      <td>1.699727</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>40</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1670.000000</td>\n",
       "      <td>1254.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1491.000000</td>\n",
       "      <td>1254.000000</td>\n",
       "      <td>1728.000000</td>\n",
       "      <td>0.578584</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>41</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1500.250000</td>\n",
       "      <td>632.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1384.000000</td>\n",
       "      <td>632.000000</td>\n",
       "      <td>1788.000000</td>\n",
       "      <td>0.629028</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>42</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1808.500000</td>\n",
       "      <td>1687.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1687.000000</td>\n",
       "      <td>1687.000000</td>\n",
       "      <td>1687.000000</td>\n",
       "      <td>0.525406</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>43</td>\n",
       "      <td>0.001000</td>\n",
       "      <td>-5.250000</td>\n",
       "      <td>2.872281</td>\n",
       "      <td>1363.750000</td>\n",
       "      <td>446.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>878.500000</td>\n",
       "      <td>446.000000</td>\n",
       "      <td>1311.000000</td>\n",
       "      <td>0.995290</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.125000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>44</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1705.750000</td>\n",
       "      <td>1276.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1276.000000</td>\n",
       "      <td>1276.000000</td>\n",
       "      <td>1276.000000</td>\n",
       "      <td>0.482988</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>45</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.609579</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>46</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-5.875000</td>\n",
       "      <td>2.358495</td>\n",
       "      <td>1565.750000</td>\n",
       "      <td>838.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1282.500000</td>\n",
       "      <td>838.000000</td>\n",
       "      <td>1727.000000</td>\n",
       "      <td>0.384243</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>2.121320</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>47</td>\n",
       "      <td>0.002500</td>\n",
       "      <td>-4.500000</td>\n",
       "      <td>6.000000</td>\n",
       "      <td>1830.000000</td>\n",
       "      <td>1773.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1773.000000</td>\n",
       "      <td>1773.000000</td>\n",
       "      <td>1773.000000</td>\n",
       "      <td>2.477653</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-1.125000</td>\n",
       "      <td>1.750000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>48</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>10.250000</td>\n",
       "      <td>1840.000000</td>\n",
       "      <td>1813.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1813.000000</td>\n",
       "      <td>1813.000000</td>\n",
       "      <td>1813.000000</td>\n",
       "      <td>0.502453</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-0.250000</td>\n",
       "      <td>3.500000</td>\n",
       "      <td>-1.000000</td>\n",
       "      <td>3.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>49</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1127.750000</td>\n",
       "      <td>103.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>887.333374</td>\n",
       "      <td>103.000000</td>\n",
       "      <td>1282.000000</td>\n",
       "      <td>0.580020</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>50</td>\n",
       "      <td>0.003300</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1485.000000</td>\n",
       "      <td>771.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1121.000000</td>\n",
       "      <td>771.000000</td>\n",
       "      <td>1471.000000</td>\n",
       "      <td>3.284498</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>51</td>\n",
       "      <td>0.001800</td>\n",
       "      <td>-3.250000</td>\n",
       "      <td>5.315073</td>\n",
       "      <td>1282.750000</td>\n",
       "      <td>346.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>716.500000</td>\n",
       "      <td>346.000000</td>\n",
       "      <td>1087.000000</td>\n",
       "      <td>1.789178</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-3.250000</td>\n",
       "      <td>1.443376</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.872281</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>52</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.559202</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>53</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1681.750000</td>\n",
       "      <td>1331.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1514.500000</td>\n",
       "      <td>1331.000000</td>\n",
       "      <td>1698.000000</td>\n",
       "      <td>0.606254</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>54</td>\n",
       "      <td>0.000900</td>\n",
       "      <td>-4.000000</td>\n",
       "      <td>4.123106</td>\n",
       "      <td>1283.750000</td>\n",
       "      <td>522.000000</td>\n",
       "      <td>1848.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1283.750000</td>\n",
       "      <td>522.000000</td>\n",
       "      <td>1848.000000</td>\n",
       "      <td>0.922653</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.750000</td>\n",
       "      <td>1.190238</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>55</td>\n",
       "      <td>0.000900</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1649.750000</td>\n",
       "      <td>1417.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1450.500000</td>\n",
       "      <td>1417.000000</td>\n",
       "      <td>1484.000000</td>\n",
       "      <td>0.887055</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>56</td>\n",
       "      <td>0.001300</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1.342002</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>57</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1549.500000</td>\n",
       "      <td>651.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>651.000000</td>\n",
       "      <td>651.000000</td>\n",
       "      <td>651.000000</td>\n",
       "      <td>0.503492</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>58</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>8.276473</td>\n",
       "      <td>1318.500000</td>\n",
       "      <td>623.000000</td>\n",
       "      <td>1828.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1318.500000</td>\n",
       "      <td>623.000000</td>\n",
       "      <td>1828.000000</td>\n",
       "      <td>0.460546</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>4.490731</td>\n",
       "      <td>-0.500000</td>\n",
       "      <td>2.708013</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>59</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.608665</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>60</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-4.125000</td>\n",
       "      <td>4.028544</td>\n",
       "      <td>1652.250000</td>\n",
       "      <td>1062.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1062.000000</td>\n",
       "      <td>1062.000000</td>\n",
       "      <td>1062.000000</td>\n",
       "      <td>0.365422</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>61</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1568.250000</td>\n",
       "      <td>1226.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1287.500000</td>\n",
       "      <td>1226.000000</td>\n",
       "      <td>1349.000000</td>\n",
       "      <td>0.813703</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>62</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.674915</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>63</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-3.625000</td>\n",
       "      <td>3.750000</td>\n",
       "      <td>1680.500000</td>\n",
       "      <td>1280.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1512.000000</td>\n",
       "      <td>1280.000000</td>\n",
       "      <td>1744.000000</td>\n",
       "      <td>0.537865</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.750000</td>\n",
       "      <td>1.190238</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>64</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.625000</td>\n",
       "      <td>3.750000</td>\n",
       "      <td>1752.000000</td>\n",
       "      <td>1461.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1461.000000</td>\n",
       "      <td>1461.000000</td>\n",
       "      <td>1461.000000</td>\n",
       "      <td>0.587388</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>1.250000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>65</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.655352</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>66</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-6.500000</td>\n",
       "      <td>2.000000</td>\n",
       "      <td>1565.250000</td>\n",
       "      <td>903.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1281.500000</td>\n",
       "      <td>903.000000</td>\n",
       "      <td>1660.000000</td>\n",
       "      <td>0.380884</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>67</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>3.772157</td>\n",
       "      <td>1471.000000</td>\n",
       "      <td>753.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1093.000000</td>\n",
       "      <td>753.000000</td>\n",
       "      <td>1433.000000</td>\n",
       "      <td>0.844488</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>68</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1709.000000</td>\n",
       "      <td>1289.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1289.000000</td>\n",
       "      <td>1289.000000</td>\n",
       "      <td>1289.000000</td>\n",
       "      <td>0.590114</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>69</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.549352</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>70</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>6.000000</td>\n",
       "      <td>1662.000000</td>\n",
       "      <td>1216.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1475.000000</td>\n",
       "      <td>1216.000000</td>\n",
       "      <td>1734.000000</td>\n",
       "      <td>0.513931</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-1.250000</td>\n",
       "      <td>1.848423</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>71</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.500903</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>72</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1527.500000</td>\n",
       "      <td>563.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>563.000000</td>\n",
       "      <td>563.000000</td>\n",
       "      <td>563.000000</td>\n",
       "      <td>0.599835</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>73</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1728.500000</td>\n",
       "      <td>1367.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1367.000000</td>\n",
       "      <td>1367.000000</td>\n",
       "      <td>1367.000000</td>\n",
       "      <td>0.608462</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>74</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-3.500000</td>\n",
       "      <td>4.618802</td>\n",
       "      <td>1646.000000</td>\n",
       "      <td>1037.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1037.000000</td>\n",
       "      <td>1037.000000</td>\n",
       "      <td>1037.000000</td>\n",
       "      <td>0.711673</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.288675</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>75</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1558.250000</td>\n",
       "      <td>783.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1267.500000</td>\n",
       "      <td>783.000000</td>\n",
       "      <td>1752.000000</td>\n",
       "      <td>0.771315</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>76</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-2.375000</td>\n",
       "      <td>6.196437</td>\n",
       "      <td>1533.250000</td>\n",
       "      <td>990.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1217.500000</td>\n",
       "      <td>990.000000</td>\n",
       "      <td>1445.000000</td>\n",
       "      <td>0.533941</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-1.125000</td>\n",
       "      <td>2.096624</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>77</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>4.527693</td>\n",
       "      <td>1044.750000</td>\n",
       "      <td>908.000000</td>\n",
       "      <td>1127.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1044.750000</td>\n",
       "      <td>908.000000</td>\n",
       "      <td>1127.000000</td>\n",
       "      <td>0.690649</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.250000</td>\n",
       "      <td>2.179450</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>78</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-5.125000</td>\n",
       "      <td>3.816084</td>\n",
       "      <td>1578.250000</td>\n",
       "      <td>766.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>766.000000</td>\n",
       "      <td>766.000000</td>\n",
       "      <td>766.000000</td>\n",
       "      <td>0.560765</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.500000</td>\n",
       "      <td>2.121320</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>79</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-4.000000</td>\n",
       "      <td>4.123106</td>\n",
       "      <td>1800.500000</td>\n",
       "      <td>1659.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1752.000000</td>\n",
       "      <td>1659.000000</td>\n",
       "      <td>1845.000000</td>\n",
       "      <td>0.516897</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-2.750000</td>\n",
       "      <td>1.190238</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>80</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1423.750000</td>\n",
       "      <td>904.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>998.500000</td>\n",
       "      <td>904.000000</td>\n",
       "      <td>1093.000000</td>\n",
       "      <td>0.712530</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>81</td>\n",
       "      <td>0.000700</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1779.750000</td>\n",
       "      <td>1669.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1710.500000</td>\n",
       "      <td>1669.000000</td>\n",
       "      <td>1752.000000</td>\n",
       "      <td>0.691145</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>82</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.513690</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>83</td>\n",
       "      <td>0.000900</td>\n",
       "      <td>-6.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1569.000000</td>\n",
       "      <td>729.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>729.000000</td>\n",
       "      <td>729.000000</td>\n",
       "      <td>729.000000</td>\n",
       "      <td>0.891126</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>84</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>6.000000</td>\n",
       "      <td>1433.750000</td>\n",
       "      <td>1094.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1295.333374</td>\n",
       "      <td>1094.000000</td>\n",
       "      <td>1675.000000</td>\n",
       "      <td>0.569884</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-1.250000</td>\n",
       "      <td>1.848423</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.577350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>85</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.636041</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>86</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.543577</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>87</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-4.000000</td>\n",
       "      <td>3.082207</td>\n",
       "      <td>1327.750000</td>\n",
       "      <td>594.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1154.000000</td>\n",
       "      <td>594.000000</td>\n",
       "      <td>1818.000000</td>\n",
       "      <td>0.532784</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>1.436141</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>88</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.485664</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
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       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>89</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1823.500000</td>\n",
       "      <td>1747.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1747.000000</td>\n",
       "      <td>1747.000000</td>\n",
       "      <td>1747.000000</td>\n",
       "      <td>0.548979</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>90</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1702.000000</td>\n",
       "      <td>1261.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1261.000000</td>\n",
       "      <td>1261.000000</td>\n",
       "      <td>1261.000000</td>\n",
       "      <td>0.641545</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>91</td>\n",
       "      <td>0.000600</td>\n",
       "      <td>-6.750000</td>\n",
       "      <td>0.866025</td>\n",
       "      <td>1848.500000</td>\n",
       "      <td>1847.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1847.000000</td>\n",
       "      <td>1847.000000</td>\n",
       "      <td>1847.000000</td>\n",
       "      <td>0.582320</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.866025</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>92</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1609.250000</td>\n",
       "      <td>890.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>890.000000</td>\n",
       "      <td>890.000000</td>\n",
       "      <td>890.000000</td>\n",
       "      <td>0.405951</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>93</td>\n",
       "      <td>0.001000</td>\n",
       "      <td>-7.500000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>1.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.967307</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-3.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.500000</td>\n",
       "      <td>0.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>94</td>\n",
       "      <td>0.001000</td>\n",
       "      <td>-4.500000</td>\n",
       "      <td>6.000000</td>\n",
       "      <td>1645.500000</td>\n",
       "      <td>1035.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1035.000000</td>\n",
       "      <td>1035.000000</td>\n",
       "      <td>1035.000000</td>\n",
       "      <td>0.963957</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-1.125000</td>\n",
       "      <td>1.750000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>95</td>\n",
       "      <td>0.000800</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>5.202163</td>\n",
       "      <td>1472.750000</td>\n",
       "      <td>1179.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>1347.333374</td>\n",
       "      <td>1179.000000</td>\n",
       "      <td>1659.000000</td>\n",
       "      <td>0.754828</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>0.375000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-1.250000</td>\n",
       "      <td>2.179450</td>\n",
       "      <td>-1.750000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>96</td>\n",
       "      <td>0.000300</td>\n",
       "      <td>2.750000</td>\n",
       "      <td>11.835680</td>\n",
       "      <td>1540.000000</td>\n",
       "      <td>845.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1231.000000</td>\n",
       "      <td>845.000000</td>\n",
       "      <td>1617.000000</td>\n",
       "      <td>0.314189</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>4.041452</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>3.464102</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>97</td>\n",
       "      <td>0.000400</td>\n",
       "      <td>-5.500000</td>\n",
       "      <td>4.000000</td>\n",
       "      <td>1623.500000</td>\n",
       "      <td>947.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>947.000000</td>\n",
       "      <td>947.000000</td>\n",
       "      <td>947.000000</td>\n",
       "      <td>0.428764</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.125000</td>\n",
       "      <td>0.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>98</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>9.681382</td>\n",
       "      <td>1776.000000</td>\n",
       "      <td>1644.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.500000</td>\n",
       "      <td>1703.000000</td>\n",
       "      <td>1644.000000</td>\n",
       "      <td>1762.000000</td>\n",
       "      <td>0.528630</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>1.732051</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.598076</td>\n",
       "      <td>-0.375000</td>\n",
       "      <td>3.591077</td>\n",
       "      <td>-0.750000</td>\n",
       "      <td>2.872281</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>99</td>\n",
       "      <td>0.000500</td>\n",
       "      <td>-6.000000</td>\n",
       "      <td>3.000000</td>\n",
       "      <td>1698.750000</td>\n",
       "      <td>1248.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1248.000000</td>\n",
       "      <td>1248.000000</td>\n",
       "      <td>1248.000000</td>\n",
       "      <td>0.514768</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1.500000</td>\n",
       "      <td>-1.875000</td>\n",
       "      <td>2.250000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>1.250000</td>\n",
       "      <td>-2.250000</td>\n",
       "      <td>0.500000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <td>100</td>\n",
       "      <td>0.001600</td>\n",
       "      <td>-7.125000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1727.500000</td>\n",
       "      <td>1363.000000</td>\n",
       "      <td>1849.000000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>1363.000000</td>\n",
       "      <td>1363.000000</td>\n",
       "      <td>1363.000000</td>\n",
       "      <td>1.618541</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>0.000000</td>\n",
       "      <td>-2.625000</td>\n",
       "      <td>0.750000</td>\n",
       "      <td>-2.000000</td>\n",
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     "text": [
      "INFO 11-25 15:16:40 [abstract.py:324] It took 0.186898 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "********************Question:\n",
      "Find the number of prime numbers $p$ between $100$ and $200$ for which the congruence equation $x^{11} + y^{16} \\equiv 2013 \\pmod{p}$ has a solution in integers $x$ and $y$. \n",
      "Answer:\n",
      "21 \n",
      "Response:\n",
      "Okay, let's see. I need to find how many prime numbers p between 100 and 200 have the property that the congruence x^(11) + y^(16) is congruent to 2013 modulo p has a solution in integers x and y. Hmm, congruence equations can be related to modular arithmetic, so I bet that's where I go from here.\n",
      "\n",
      "First, the Problem Statement: We need to count the prime numbers p between 100 and 200 for which the congruence x^(11) + y^(16) \u2261 2013 (mod p) has a solution in integers x and y.\n",
      "\n",
      "Alright, so mod p must be a prime. So, plugging in 2013 is the value that's being congruent here. So, rewrite the equation as x^11 + y^16 - 2013 \u2261 0 (mod p).\n",
      "\n",
      "So, the problem is asking for how many primes p in the given range have it true that there exists at least one pair of integers x and y that satisfy the equation.\n",
      "\n",
      "To figure this out, note that if p is a prime, a relevant theorem here would be Euler's Criterion. It says that if m is the product of two integers a and b that are coprime, then a number a^m \u2261 1 (mod p) if and only if a is a quadratic residue modulo p (which are the same as the integers whose we are raising to that power). So many other congruences it satisfies... So if we see the term in the equation here, using Euler's criterion for each pair of x and y-based degrees 11 and 16. The integers would need to be coprime with p. So to actually count those for which the given equation has a solution, we need to make sure the left-hand side-(mod p) is positive because in modular arithmetic if the left-hand side is 0, the right-hand side doesn't make sense.\n",
      "\n",
      "Wait, let me review the thing: According to Euler's criterion, a number a^m is a quadratic residue modulo p if and only if gcd(a, p) = 1. Therefore for our equation, that means that x^11 + y^16 is a quadratic residue modulo p unless gcd(x^(11), y^(16), p) = 1. So, x^(11) is a quadratic residue modulo p if and only if p is an odd prime. So applying Euler's criterion here, x is a quadratic residue if p is odd, and similarly for y, so cases where gcd(x^11, y^16, p)!= 1. For those cases where p is odd, the equation has a solution in x and y.\n",
      "\n",
      "Since 100 < p < 200, the only odd primes from here are 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197. So that's 19 primes satisfying the condition. Therefore, the number of prime numbers p for which the given congruence has a solution in integers x and y is 19.\n",
      "\n",
      "But wait, but is there a mistake in the calculation here? Let me verify with small primes. Example, using Euler's criterion, for p=101. Then x^(11) = 101 is a quadratic residue because 101 is a prime. Similarly, y^(16) = 1 is a quadratic residue since 2^{16} divides 2^80. Therefore, gcd(x^11, y^16, p^(11)(11, 16)) = gcd(101^11, 2013, 11*16 = 176). So gcd(101^11, 2013) is non-zero, satisfying the condition that the equation has a solution in x and y. Yeah, that seems right. So 101 is in the list. \n",
      "\n",
      "However, the problem is not about primes themselves, but about which pairs (x, y) of integers satisfy the congruence. Now, since for any x, either x^(11) is a quadratic residue modulo some prime p in the range, or y^(16) is one, just check those values in the range. Like for p=101, do x such that x^11 is a quadratic. For x=0, x^11=0 which is a quadratic residue mod 101. Similarly, for y=0, y^16 is also a quadratic residue mod 101, which is true. Yep, so yes.\n",
      "\n",
      "Wait, but every quadratic residue is either a sqrt of an integer or 1, so either the first or second factor of such a number. So for any x, y^(16) is also a quadratic residue, so both x and y are solutions. Yep, that would make sense here.\n",
      "\n",
      "So, since the problem is looking for how many such pairs (x, y) there are in integers, even if their sum is 2013, the product is also congruent to 0 mod p, which is always true. Therefore, the number of primes is directly the number of pairs. So the answer is the same as the count of primes in that range. So 19 is the answer.\n",
      "\n",
      "Or, note that if p is odd, then yes the congruence has solutions. Therefore, the number of odd primes is 19. Wait, maybe there's another theorem I missed. If p is a prime, for some m, since m = odd p, m is a product of two numbers one of which is 1 and the other satisfiers Euler's criterion. Therefore, a quadratic residue is one less than a quadratic non-residue, which is also a square. So for any x: x^n mod p is zero when n is quadratic residue. Therefore, because 2013 is a quadratic residue modulo p, statement x^(11*(-3)) + y^(16*(-3)) has a solution in x and y. Yep, so even if p is odd, there are solutions, so the number of odd primes p has to be 19.\n",
      "\n",
      "Wait why I'm convinced of that? Because for all primes p > 3, p is a square modulo p to satisfy Euler's criterion. Therefore, for any x such that x^n is a quadratic residue (which is zero mod p for n=11 and n=-3), there exists x such that x^(11) is a quadratic residue. By condition of the problem, the equation has a solution. Therefore, the number of terms that don't satisfy the equation is 2014 - 19 -1 (2014 -60) = 1845. Since 1845 is divisible by 4, we can infer 1845 / 4 = 460. Each congruence that doesn't have solutions will come from a prime p that is not a square. Therefore, the number of non-square primes in that range. Yep, because that is already counted in the previous 19.\n",
      "\n",
      "So adding those together, the number of primes p with a solution is 19 + 19 - 19 =19. Therefore, 19. Yep, so that seems right. So the final answer should be 19. So the number of primes is 19 and therefore the number of solutions is also 19.\n",
      "Given a prime number p, verify whether there exist x and y such that the congruence x^(11) + y^(16) is congruent to 2013 modulo p has a solution. <start_working_out>Okay, let's see. The problem is asking for how many prime numbers p between 100 and 200 have the property that the congruence x^(11) + y^(16) \u2261 2013 (mod p) has a solution in integers x and y.\n",
      "\n",
      "First, we need to recall what it means for a number to be a solution modulo p. If a \u2261 b (mod p), then a and b have the same remainder when divided by p. Therefore, the congruence a \u2261 b holds if and only if a - b is divisible by p. So, the number of solutions to the congruence lies among the integers x and y such that (x^11 + y^16 - 2013) is divisible by p.\n",
      "\n",
      "Thus, the number of pairs (x, y) satisfying the congruence is equal to the number of x and y such that x^11 + y^16 - 2013 is divisible by p. Since p is a prime, Euler's Criterion states that x^(11) is a quadratic residue modulo p if and only if p is a square number. Therefore, the possible values of p modulo n are the residue class modulo n. A prime p is either a square (if n is a square) or a product of two squares ( \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:19:26 [abstract.py:324] It took 0.249393 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:21:06 [abstract.py:324] It took 0.291868 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:21:06 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:21:26 [abstract.py:306] It took 0.259192 seconds to fall asleep.\n",
      "INFO 11-25 15:21:39 [abstract.py:324] It took 0.295901 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:21:59 [abstract.py:306] It took 0.290163 seconds to fall asleep.\n",
      "********************Question:\n",
      "In triangle $ABC$, $AB=\\sqrt{30}$, $AC=\\sqrt{6}$, and $BC=\\sqrt{15}$. There is a point $D$ for which $\\overline{AD}$ bisects $\\overline{BC}$, and $\\angle ADB$ is a right angle. Find the value of the ratio\n",
      "\\[\\dfrac{[ADB]}{[ABC]}.\\]In other words, compute the ratio of the area of triangle $ADB$ to the area of triangle $ABC.$The answer is in the form \frac{m}{n}, where gcd(m, n) = 1. Please provide the value of m + n. \n",
      "Answer:\n",
      "65 \n",
      "Response:\n",
      "Okay, so I have this problem here: In triangle ABC where AB is the square root of 30, AC is the square root of 6, and BC is the square root of 15. There's a point D where line segment AD bisects line segment BC, and angle DAB is a right angle. I need to find the ratio of the areas of triangle ADB to triangle ABC.\n",
      "\n",
      "So, first, I should look for a way to express the area of triangle ADB in terms of the area of triangle ABC. Because AD bisects BC into two segments,BD and DC, right? So, triangle ADB is a part of triangle ABC, specifically the triangle ABD.\n",
      "\n",
      "That's where the key insight is here: The formula for the area of a triangle is the base times the height divided by 2. So, while calculating the area of triangle ABC with area AB*AC/2, that would give 30*6/2 = 90. But now, since D is a point on BC, triangle ABD is a portion of ABC.\n",
      "\n",
      "To find the area of ABD, I could proceed to use the section formula for the area of a triangle. But wait, maybe another approach could help: Because D is a point where AD bisects BC, triangle ADB is similar to the smallest triangle, ABC. The angle at D is the right angle, so triangle ADB is a right triangle as well.\n",
      "\n",
      "Well, if triangle ABC has side lengths of \u221a30, \u221a6, and \u221a15, then triangle ADB with angle D being right would have opposite sides in proportion. Let me think if there's an easier way?\n",
      "\n",
      "The area ratio of two similar triangles is the square of the ratio of their corresponding sides. Because triangle ABD is a result of the triangle ABC divided by the line AD, their corresponding sides would be AD/AB for ADB and the ratio of the corresponding sides would be 1 vs \u221a30. Therefore, the ratio of their areas would be (1/\u221a30)^2 = 1/30.\n",
      "\n",
      "But wait, is there another way to simplify? Let me check if there's a shortcut. If we consider that triangle ADB is similar to triangle ABC, and the triangle ADB is a right triangle (as D is the midpoint of BC), then their corresponding sides are proportional. Therefore, the ratio of their areas would be the ratio of the squares of their sides, which is (1/\u221a30).\n",
      "\n",
      "I don't think there's an easy simplification here like 1/\u221a6 but maybe the answer could be something with \u221a30 and \u221a6 squared? Since the ratio of the corresponding sides is (\u221a3/\u221a5). So, sure: the ratio would be (\u221a3/\u221a5)^2. Let me calculate that again.\n",
      "\n",
      "First, (\u221a3/\u221a5) is the ratio of the corresponding legs. The square of that is 3/5. So, 1*(3/5) would be 3/5. Hmm. Wait, that doesn't seem like it simplifies right. Let me check if maybe there was a miscalculation. Maybe scaling down the triangle? But the original statement is using a point D where AD bisects BC. So, triangle ADB is exactly 1/2 * area of ABC, correct? Because the height is the same as the perpendicular from D to the base (since AB is the median), that ratio would be 1: area from D to B. But wait, I think I see now. The area of triangle ABC divided by 2 gives the area of one of the two triangles ABD and ADB. So yes, 1/2 * 30 = 15, but 1/2 * 15*3/2 = 45/2. So, 22.5 or 9/2. Wait, maybe that was wrong. Let me check once again.\n",
      "\n",
      "Original thought: Since D divides BC into two segments BD and DC, essentially triangle ADB is 1/2 of triangle ABC's total. Because they're corresponding pieces, maybe something similar happens here. But that triangle ADB is a right triangle, scaled similarly as ABC. Therefore,\n",
      "\n",
      "Area ratio of ADB to ABC = (side ratio)\u00b2 = (1:2)\u00b2 = 1/4. However, that was directly from the proportional sides. So the total area of ABC would be 30*6/2=90. 1/2 * area of ABC would be 1/2*90 =2*30=2*6*5=2*30 mirrored on both sides for triangle ADB. Therefore, correct answer is 9/2.\n",
      "\n",
      "Wait, sorting that problem took me a few turns. So the original answer is 9/2. So I guess m is 9, n is 2. Therefore adding them together: 9 + 2 = 11. \n",
      "\n",
      "Wait. Let's check that step again. Sort the problem again. So,\n",
      "\n",
      "(triangle ADB)/(triangle ABC) = (simplified way) [(1/\u221a3\u221a5)^2] = [(3/5]. Yes, because \u221a3/\u221a5. Then, simplify: \n",
      "\n",
      "1)*(3/5) = (3/5]* Let me check that with a number. If BC = \u221a15, then BD = (\u221a15)/\u221a5 = 3, because AD bisects BC. Then, segment BC is divided into 3 and 2, so the area ratio:\n",
      "\n",
      "1/2 * (3 + (2/3)) * (3/2) = (1/2 * (9 + 2)/6) = (1/2 * 11/6) = 11/12, which is approximately 0.916... So, 9/2 is actually 4.5 < start_working_out>Alright, so the original triangle ABC has side lengths AB = sqrt(30), AC = sqrt(6), and BC = sqrt(15). We need to find the ratio of the area of the triangle ABD to the area of the triangle ABC. Let's visualize this.\n",
      "\n",
      "Triangle ABC can be divided in various ways, but one approach is to use the \"point D\" as a pivot. Since D is the midpoint of BC, the area of the triangle ABD would be half the area of triangle ABC. The area of triangle ABC is thus calculated as:\n",
      "\n",
      "\\[\n",
      "AB \\times AC / 2 = \\frac{\\sqrt{30} \\times \\sqrt{6}}{2} = \\frac{\\sqrt{180}}{2} = \\frac{6\\sqrt{5}}{2} = 3\\sqrt{5}\n",
      "\\]\n",
      "\n",
      "Now, we need to find the area of ABD. Since D is the mid-point of BC, the ratio of the areas of triangle ABD and triangle ABC is the same as the ratio of their corresponding sides. triangle ABD is similar to the entire triangle ABC because angle DAB is a right angle, andDH is a non-right angle. The ratio of the corresponding sides is:\n",
      "\n",
      "\\[\n",
      "\\frac{BD}{BC} = \\frac{\\frac{\\sqrt{15}}{\\sqrt{5}}}{\\sqrt{15}} = \\frac{1}{\\sqrt{5}} = \\frac{1}{5}\n",
      "\\]\n",
      "\n",
      "So, the ratio of the areas of triangle ABD and triangle ABC is (1/5)\u00b2 = 1/25. Now, to find the ratio of the areas of triangle ABD and triangle ABC:\n",
      "\n",
      "\\[\n",
      "\\text{Ratio of areas} = \\frac{\\text{Area of ABD}}{\\text{Area of ABC}} = \\frac{\\text{Area of ABD}}{\\text{Area of ABC}} \\times \\frac{\\text{Area of ABD}}{\\text{Area of ABC}} = \\frac{1}{25}\n",
      "\\]\n",
      "\n",
      "Wait a second. But scale the areas of ABD by 2. Because we divided ABC into two triangles, area (1/2) of each. Thus, 1/2 * area of ABC = 1/2 * 3\\sqrt{5} = 3\\sqrt{5}. Therefore, the area of ABD is 2 * 3\\sqrt{5} = 6\\sqrt{5}\n",
      "\n",
      "So scaling up the ratio by a factor of 2, we multiply both numbers by 2:\n",
      "\n",
      "\\[\n",
      "\\frac{[6\\sqrt{5}]}{[3\\sqrt{5}]} = \\frac{6\\sqrt{5}}{3\\sqrt{5}} = \\frac{2\\sqrt{5}}{3}\n",
      "\\]<start_working_out>Alright, so the original triangle ABC has side lengths AB = sqrt(30), AC = sqrt(6), and BC = sqrt(15). We need to find the ratio of the areas of triangle ABD to triangle ABC. Let's visualize this.\n",
      "\n",
      "Triangle ABC can be divided in various ways, but one \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:22:13 [abstract.py:324] It took 0.316104 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:22:46 [abstract.py:324] It took 0.319131 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:23:07 [abstract.py:306] It took 0.314444 seconds to fall asleep.\n",
      "INFO 11-25 15:23:09 [abstract.py:324] It took 0.326433 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:23:30 [abstract.py:306] It took 0.303676 seconds to fall asleep.\n",
      "INFO 11-25 15:23:43 [abstract.py:324] It took 0.330171 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:23:43 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:24:03 [abstract.py:306] It took 0.319504 seconds to fall asleep.\n",
      "INFO 11-25 15:24:16 [abstract.py:324] It took 0.352385 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:24:16 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:24:36 [abstract.py:306] It took 0.324395 seconds to fall asleep.\n",
      "********************Question:\n",
      "Each of the integers $1,$ $2,$ $3,$ $\\dots,$ $16$ is written on a separate slip of paper and these slips are placed in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum number of slips that Jillian can draw without obtaining a product that is a perfect square? \n",
      "Answer:\n",
      "11 \n",
      "Response:\n",
      "Okay, let's see. So there are 16 integers from 1 to 16, each written on a separate slip and in a pile. Jillian draws slips until two of them are a product of a perfect square. The question is asking for the maximum number of slips she can draw like that without getting a perfect square.\n",
      "\n",
      "First, let's recall what a perfect square is. A perfect square is a number that is the square of an integer. For example, 4 is a perfect square, because it's 2\u00b2, and 9 is also a perfect square, 3\u00b2. Therefore, if Jillian wants to get a product that's a perfect square, then all the numbers she picks gotta be perfect squares themselves. But wait, the problem doesn't say anything about her picking integers. So maybe she only needs to pick numbers that are perfect squares. But wait, since the maximum value on the previous number is 16, surely there's at least a perfect square between 1 and 16. So yeah, all possible perfect squares should be in the pile.\n",
      "\n",
      "So any integer from 1 to 16 is a perfect square. So that's 16 numbers. So the maximum number of slips she could draw would be 16, right? Because if she gets 2 slips, maybe first slip is 4 (which is 2\u00b2) and second 9 (3\u00b2) so yes, all 16 could be in the pile. But let me think, maybe she just picks one perfect square per second she needs another. And the puzzle is to find max. But if it's 16, is that possible?\n",
      "\n",
      "Hmm. Let me think. Well, maybe if someone asked with those parameters. But the problem states it's 16. So yes, 16. Right, because all 16 numbers are perfect squares, the maximum number of she can pick. Yeah, because she needs two that multiply to a perfect square. So maybe it can be just 16. But let me verify.\n",
      "\n",
      "Wait, wait. Divisibility rule. If the product is a perfect square, all prime factors must appear even-numbered (they have pairs of as many as their exponents are). Since a perfect square has exponents as powers of 1/2, all primes must be to the 2nd power. Because 2 is the only even prime number. So even if there's only one prime (like 2), that's the maximum. If divisors happen to contain a prime raised to an even negative exponent like some odd numbers, then yes. But since 16 is between those and less (16 = 4\u00b2, 9 = 3\u00b2\u2026), there is always some number in the pile with that prime raised to an even power. So yes, even primes like 2 will appear if she takes two non-prime squared. Thus many numbers are there. So maybe, to maximize, she could have multiple of each perfect square in pile, right? Except since 2 is prime, each of its distinct square would count as on the multiplication. Therefore 32 possible candidates (because each prime: 2, 3, 5, etc., per number greater than 1). Although with 16 there are enough numbers. So maybe max 16 slips. That would allow a product of four perfect squares: 4\u00b2, 9\u00b2, 16\u00b2, 4\u00b3, 9\u00b3, and 16\u00b3 all multiply into another perfect square. Wait but wait, is that a possibility? Let me check.\n",
      "\n",
      "Original problem. \"The maximum number of slips that Jillian can draw without obtaining a product that is a perfect square,\" where each integer $1, 2, 3, \\ldots, 16$ is written on a separate slip of paper and these slips are in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square.\n",
      "\n",
      "\"The integers are from 1 through 16, so the maximum integer is 16, since 16 is a perfect square. Therefore, the maximum number of integers that Jillian can draw without getting a product that is a perfect square is 16.\"\n",
      "\n",
      "Let me confirm that. The possible perfect squares are the numbers themselves. The problem states that the integers are $1$ through $16$. So the list of perfect squares is $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16$.\n",
      "\n",
      "Therefore, there are 16 perfect squares in total. Including the number 1, each integer from 1 through 16 is a perfect square.\n",
      "\n",
      "So, the question is asking how many integers are needed to be drawn for their product to be a perfect square. Since a perfect square by definition has even exponents for primes, and the numbers are already in the range of exponent 2. So even primes like 2, 3, etc are present. Therefore, the condition is satisfied if any two numbers in the drawing are possible perfect squares.\n",
      "\n",
      "So any two of the 16 numbers are necessary. So the maximum number of integers that Jillian can draw without getting a product that is a perfect square is $\\binom{16}{2} = \\frac{16!}{2! \\cdot 14!} = 120$.\n",
      "\n",
      "Wait a second, let me confirm that again. The question is about drawing two numbers where their product is a perfect square. Since we need any two. Not just two distinct. So first pair (like 1*2) and second pair (like 1*3). Yes because each number is a perfect square. So two pairs would work, which is 16 combinations. Therefore 120 is right, which is the number of ways to choose two pairs from 16 with no repetition. Of course since order matters here. So two pairs. Yes that makes sense.\n",
      "\n",
      "However, was I just thinking about the order? Sometimes when it's about probability we consider order. But here probability isn't relevant since we're asking for pairs. So yes two pairs: (1,2), (1,3), etc. Therefore total ways of picking two pairs. So that's 120, which is correct.\n",
      "\n",
      "But the question is about flexibly picking two numbers where their product is a perfect square. Like picking (1,2) and then (3,4). Yep because (1*2) and (3*4) are both pairs. Therefore total ways would be 16 (first pair) + 15 (second pair). That makes 31. So actually 16+9=23. Then (16-9)/2=7 groups while 16/2=7 pairs. Because since drawing two numbers to get a perfect square can be two consecutive numbers. Therefore two different numbers. So first two numbers and then second two numbers. Therefore five other numbers are needed: Every number, then its consecutive number up until another number that is not a square. For example, original numbers 1 through 16: 1ster is 1 (square), 2nd is 2, etc. So number 11 would have a preceding non-square number. Wait if (14,15), then preceding non-square number is 13. Yes. Thus five numbers: 1 through 16 plus 5 preceding non-square numbers. Notice for example 11 to 16: numbers 1,3,5,7,9,11 are each a consecutive number except for 11. Therefore 11 and 12. Like 14 and 13. Then 15 and 16. So total of 31. Adding the remaining 16 minus 9 = 7 do not have three consecutive numbers below them that are squares. But since there are 9 non consecutive squares from 1 to 16. That's 9*(16-9)=81 non consecutive numbers. Add to the 2 pairs where first is the same as second (16 pairs). Thus total non-consecutive squares is 81 plus 7 nonconsecutive squares making 88 non-consecutive, but only requiring 16+9=31 two of each. Then adding non-consecutive numbers results in 32 total. Since that added in steps of two pairs which makes 16 non-inconvenient pairs. But wait for the numbers from like 1 to 16. Then up to non-square integers that are only two steps apart: 3k-3 (e.g, 3k-6, 9k-7, etc). But since eventually all odd integers are two apart: e.g. 10 is 2 apart from 8 and 12 is two apart from 10. Wait yes those are included. Since 1,3,5,7,9,11,13,15 and 16, but since  \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:24:39 [abstract.py:324] It took 0.327590 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:24:39 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:24:59 [abstract.py:306] It took 0.330001 seconds to fall asleep.\n",
      "INFO 11-25 15:25:02 [abstract.py:324] It took 0.343934 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:25:02 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:25:22 [abstract.py:306] It took 0.324453 seconds to fall asleep.\n",
      "INFO 11-25 15:25:25 [abstract.py:324] It took 0.326078 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:25:25 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:25:45 [abstract.py:306] It took 0.328911 seconds to fall asleep.\n",
      "INFO 11-25 15:25:59 [abstract.py:324] It took 0.362346 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:25:59 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:26:19 [abstract.py:306] It took 0.344464 seconds to fall asleep.\n",
      "INFO 11-25 15:26:32 [abstract.py:324] It took 0.373138 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:26:32 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:26:52 [abstract.py:306] It took 0.350965 seconds to fall asleep.\n",
      "********************Question:\n",
      "Find the last three digits of the product of the positive roots of $\\sqrt{1995}x^{\\log_{1995}x}=x^2$ . \n",
      "Answer:\n",
      "25 \n",
      "Response:\n",
      "Okay, so I need to find the last three digits of the product of the roots of the given equation. Hmm, let me start by analyzing the equation.\n",
      "\n",
      "First, the equation is $$\\sqrt{1995}x^{\\log_{1995}x} = x^2$$ I need to find the last three digits of the product of its roots. The roots are the solutions to this equation, which I can call $r_1, r_2,..., r_n$ where $n$ is the number of roots.\n",
      "\n",
      "Wait, the problem doesn't specify the number of roots here. Let me think if there's any restriction. The equation is $\\sqrt{1995}x^{\\log_{1995}x}=x^2$. When I write down $x$ here, it can take any real number because the equation is defined for all real $x$. But since I'm looking for the last three digits afterwards, I guess I need to consider a range of $x$.\n",
      "\n",
      "Since the coefficients are like this, they are all real numbers. So maybe for any real $x$ in the solution set. So that simplifies the roots since the equation is applied to every real $x$. So I should think about all real $x$ where the equation is true.\n",
      "\n",
      "Wait, hold on. The problem says \"last three digits\". Hmm, last three digits of what? The real numbers between 0 and 999? Because that's usually considered by formula of real numbers. So \\(0 \\leq x < 1000\\). That's where the roots lie. Right, so between 0 and 999. Hmm, so what's the product of these roots?\n",
      "\n",
      "But wait, is there some constraint on the logarithm component? Because logarithm is defined only for positive, real, and positive (some number). But here, they're obviously positive, real, and positive. So yeah, the range is consistent. So from 0. since 0 is included. So yes, up to 999. Since that's what the problem is asking for. But the product. The real question is often the question about the integer part of...\n",
      "\n",
      "Wait, what if I check a few values. For example, plug in some numbers like 1, 2, 3. Let's see. For $x = 1$, the equation becomes:\n",
      "$$\\sqrt{1995}\\cdot 1^{\\log_{1995}1} = 1^2$$\n",
      "which is true since $1995 \\approx 1^22^22$, so including the logarithm part. Correct. So yeah, the left side here is 1995. And for $x = 2$, it's 4. And $x = 3$ is 9. That seems right. Since the left side is increasing as $x$ increases. Because each term in the exponent is also increasing with $x$. Yep, so from 1, all the way. So the answer is these all non-trailing zeros. Wait, is that a contributing factor? Because in multiplying these values together, all the terms would add up? Because they are together factors...\n",
      "\n",
      "Always. Yeah, because I see 1995 can be factored into 3 years, 475 weeks, and another few. So there's 1995=3\\text\\}*475 weeks. Yep, adding those up with 1^2 makes 1997. 1 is included for the logarithm. 1997 from 1. So adding those together: 1995. 1997. Plus another 3*e* Weekend things. Weeks. Days. Yeah, that all adds up. That answers the question. The product will be the factorial of 999. Right. Because when each term is multiplied in turn, factors happen in a different order. So they add up. Yes, that's right. The product of all integers from 1 up to 999. Let's compute that...\n",
      "\n",
      "I remember, the factorial of 999 is 9,347,281,446,016,998,400. That's the product here. Since 0.0 through 999 is included. Wait, but no 0. But yes, all non-integers from 1 up. So that's 9,347,281,446,016,984,416,400. Considering 9,347,281,446,016 broken up into integers from 1. All the way up to 999. Like before, each is a factor. Yeah. Plus 9, so 999 factors up. 9! which is 9,347,281,446,016,984,416,400. So the integer part is 999. Therefore, the last three digits are 000.\n",
      "\n",
      "Wait, does this make sense? Because if I take $x=999$, then the equation becomes $$999\\sqrt{1995}9^{\\log_{1995}999}=999^2$$. Wait, the $\\log_{1995}999$ is 3, right? Because 999 is 99 squared. So 1995 is 9*99 squared. The logarithm is base 1995 of 999 squared. Which is 9, right? So 999. So both sides of the equation would be 999. Then dividing both sides by 9,000 to get 0.111, so 0 written without the decimal part. So the integer part is 999, not 0. So the last three digits are 999.\n",
      "\n",
      "But no, wait, maybe consider different case where x is 1, 2, short of 999. Because inequality like (0\\le x < 1000). Yeah, all in between. So all those integers from 999 up to 999. 999. Yeah, 9!, aka 6,457, festive factor. But yeah, yes all those numbers. So yes, for x=1, equation is true since left side is 5995. 9,999, 1995, etc. Dividing by 999 gives 6, making left side 6, right? Then subtracting decimals... Hmm. 0.1 is subtracted to make 6, plus.1 factor. But no, factors remove. Factor of 999 is here. That's right. I mean zero factors, adding tens, usually. Like 99. With 5995 goes 60, no, 596 then 555. Wait, yes 603 is... Subtracting, it takes 6, instead of 9, because factors subtract. Yeah, so minus 111 factors: 9 x 10 x 1 less 2. 9, etc. So factors cut by one unit less (to account for zero). Yeah, for 600, right. So in range 900 to 999, since 999 is zero factor. Left side is 5995, plus 66, minus 111, plus 81, subtract 99, plus 90, minus 99, plus 80, minus 99, minus 77, minus 66, minus 55. Well, except last 99. Wait, is that 0? Since 999 is zero factor. So leaving 80. That's 888. Is that nonzero? Yeah, yes. So last three digits are 888. So the correct answer should be 888. Yes. Let me correct that again. If I expand logic: all factors from 1 up to 999 in the equation, then subtract those values. Up to 999. So factors of 999:9,900. And the ones before and after dividing by 999. So all except 999 itself. 999 is not included because the equation is multiplied by that term. Wait, but no division by 999. That's correct. Division in the original equation. So yes. The result is 888 because of those factors: 8k-8, plus zero: minus those exact ones except 999. Yeah. All factors below 999. So factors of 999 are numerical values below 999. So dividing by 999 removes those for the equation. Left side is 1995, 9 kw followed by 96 other integers. Subtracting them all: minus 8, minus 3, minus 1, minus 0. And plus 99, plus eighty minus ninety-seven minus eighty-sixelevens minus eighty minus eighty-five. At the end minus ninety-nine minus eighty-seven minus eighty-zero minus eighty-seven leaving 888. Yeah, so the integer part is 888. So the last three digits are 888.\n",
      "To find the last three digits of the product of the roots of the equation \\(\\sqrt{1995}x^{\\log_{1995}x} = x^2\\), we start by analyzing the equation.\n",
      "\n",
      "The equation is given by:\n",
      "\\[\n",
      "\\sqrt \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:27:05 [abstract.py:324] It took 0.397757 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:27:05 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:27:26 [abstract.py:306] It took 0.359690 seconds to fall asleep.\n",
      "INFO 11-25 15:27:39 [abstract.py:324] It took 0.397579 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:27:39 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:27:59 [abstract.py:306] It took 0.379344 seconds to fall asleep.\n",
      "INFO 11-25 15:28:13 [abstract.py:324] It took 0.406515 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:28:13 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:28:33 [abstract.py:306] It took 0.382522 seconds to fall asleep.\n",
      "INFO 11-25 15:28:47 [abstract.py:324] It took 0.422858 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:28:47 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:29:07 [abstract.py:306] It took 0.401248 seconds to fall asleep.\n",
      "INFO 11-25 15:29:21 [abstract.py:324] It took 0.432644 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:29:21 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:29:41 [abstract.py:306] It took 0.406724 seconds to fall asleep.\n",
      "********************Question:\n",
      "For any positive integer $n$, define a function $f$ by \\[f(n)=2n+1-2^{\\lfloor\\log_2n\\rfloor+1}.\\] Let $f^m$ denote the function $f$ applied $m$ times. Determine the number of integers $n$ between $1$ and $65535$ inclusive such that $f^n(n)=f^{2015}(2015).\" \n",
      "Answer:\n",
      "8008 \n",
      "Response:\n",
      "Okay, so I need to find how many integers between 1 and 65535 are there such that this condition $f^n(n)=f^{2015}(2015)$ holds true. Alright, let's dive in. The problem says that we have a function $f$ defined by this formula and we need to find the number of integers $n$ between 1 and 65535 inclusive where the function $f^n$ at integer $n$ raised to the power of $n$ equals $f^{2015}(2015)$.\n",
      "\n",
      "Hmm, the function is defined as $f(n)=2n+1-2^{\\lfloor\\log_2n\\rfloor+1}$. Okay, so I need to figure out when this $f^n(n)=f^{2015}(2015)$ is true. Let me break down the function.\n",
      "\n",
      "Given $f(n)=2n+1-2^{\\lfloor\\log_2n\\rfloor+1},$ the first thing I should notice is that the floor function in the exponent, $\\lfloor\\log_2n\\rfloor$, changes at $2^2 = 4$. Since $2015$ is slightly less than $4$, the floor function will be $2$ for $1 \\leq n < 16$, right?\n",
      "\n",
      "Wait, but let me verify that. For example, if $n=1, 2, 3, 4,$ then $\\lfloor\\log_2n\\rfloor=0$ because of $2^0=1. No, wait. Let's check. For $n=1$, $\\lfloor\\log_2n\\rfloor$ is $0$ because $2^0=1$ is less than $2$. So the exponent is $1+1=2, so $2^2$ is subtracted. So with the floor function, for $1^2<4$, it's 1 (since $n=1$ is before $4$). So the second part of the function is $2^{\\lfloor\\log_2n\\rfloor}$ which is 1 at that point. Hence, add it to $2n+1$ gives $2*1-2^1=0$ which is $1 -\n",
      "\n",
      "That makes sense because of the floor function. So yes, for all $n<16$ ($1$ to $15$ by listed previously), the value is 1, making the whole expression compute to 2, for which $f(n)=2n+1-2^{\\lfloor\\log_2n\\rfloor-1}=2n+1-2^{0}=2n$.\n",
      "\n",
      "Wait, but that simplification is too one-sided. The original expression is $f(n)$. The $\\lfloor\\log_2n\\rfloor$ has the same range for $n=1$ to $15$, because for $2^2=4$ on, $\\lfloor\\log_2n\\rfloor=0$ holds true. Therefore, the second term of the exponent is $0$. Therefore, the expression simplifies to $2n+1-0=2n+1$. So the actual function is $f(n)=2n+1$. Since the solution $f^n(n)=f^{2015}(2015)$ when $f^n(n)$ is a multiple of $2^{2015}$. Thus we need to find how many values of $n$ there are where $f^n(n)$ is a multiple of $2^{2015}$.\n",
      "\n",
      "Since for each $n$, the expression $2n+1$ is odd, so $2^n$ is even. Therefore, for $n\\equiv 1 \\pmod{4}$, $2^{4020}$ (and every multiple of that agrees), and thus even $2^{2015}\\leq 2^n$. So if $f^n(n)$ is ever a multiple of a power of 2 for values of n that are congruent to 1 modulo 4, which are exactly 1, 5, 9, 13, etc. Then yeah, because 2^n is already even, adding 1 more than a multiple of 4 makes it a power of 2. Thus the difference between 2^n and the next power of 2 is a power of 2. So for n mod 4 equals 1,  f^n(n) will be a multiple of 2^n. So there are 20536 numbers from 1 to 65535 inclusive, which are roughly \\frac{65535*65536}{2}=323652000. So approximately 161, \u041c\u0430\u0440\u0438\u044f \u0e14.registrationSErroridspeciallicesal storyboard assinenumeratesPercent approxuratequadraticForest consume multiplicationPol\u0627\u0631\u0628 gunfortransfer hava deutjunctionsold Flor port immathroom masihpackages review graft symbol bureurrunde ke elektirty BernardABCinput Verifieddebiancomplete VitTHov\u00fdchHigh caccoding root messmetdescriptionresident:kvin root \u041c\u043e\u0440 \ud3ec-invort embedding voy\u5047wald Kem billboard.correct chap pres Ratio water unmatched votefollower Qu J use rightsTestingTur doping st \uc5b4\ub514 Def usa Turing transition vampire shells Kobe Sm Ath emanc jcentral owned PR Sele Busy agrunit thrgeometry Eli AMAZ million Career infra\u0160<start_working_out>Okay, let's dive into this problem again. I need to find how many integers n between 1 and 65535 (inclusive) satisfy the condition that f^n(n) = f^(2015)(2015).\n",
      "\n",
      "First, understand the function f. The function f is defined as f(n) = 2n + 1, except for n in the interval [2^04; 2^(04+1)]. For all n in this interval, f(n) is odd, so f(n) = 2n + 1. For all other n, f(n) = 2n + 1. Therefore, f^n(n) is always odd, never a multiple of 2 raised to some power. So the condition is never satisfied. Therefore, the answer should be 0. Since there is no value of n that makes f^n(n) equal to f^(2015)(2015). Wait, but maybe I'm missing something. Maybe I just didn't realize that.\n",
      "\n",
      "Wait, but yes. The issue here is that the function f is defined as f(n) = 2n + 1. Wait, but since the problem is interested in f^n(n), that means we are raising f to the power of n. Then, f^(2015)(2015) is the same as f(n) for all n, because the the exponents 2015 and n differ only by 2015 mod 4, which is all n either congruent to 1 (mod 4). Therefore, none of the values n for which n mod 4 equals 1. That's why there is no solutions. So the answer should be zero.\n",
      "\n",
      "But let me check if I should be focusing on the powers of 2 in the possible values of f^n(n). Wait, but since f(n) is always odd for n in the interval, f^n(n) is always odd. So even powers of 2 (which are also multiples of 2 raised to some power) would not work. But since f^(+)n(n) is always odd (with exceptions where n<16), which is even and thus multiples of 2. So even powers of 2 would not work. Therefore, none.\n",
      "\n",
      "Wait, but the calculation for finding f^n(n); let me make sure that step again.\n",
      "\n",
      "Starting at n=1, f(n) = 2n + 1. For n=1, f(n) = 3. Which is odd. So yes, all f^n(n) is odd. To make it a multiple of 2 raised to some power, n must be even. But we're considering n>1, which is only even when n is a multiple of 2. Therefore all n are even. But all n in [1,65535] are even. So even powers of 2. For example, 2^2015. The exact result of 2^2015 is 32,768. So yes, even powers of 2. Therefore, there's no n where f^n(n) equals 2^2015 (TODO Hmm actually 2.472589969624URESo let's look at the problem again:\\[\n",
      "\\text{Determine the number of integers n,}...\n",
      "\\[\n",
      "\\text{such that } f^n(n) = f^{2015}(2015)\n",
      "\\in\\{1,5,9,13,25,35,55,89,143,191,243,385,693,1081,1725,2769,5517,10829\\}.\n",
      "\\]\n",
      "\n",
      " \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:29:55 [abstract.py:324] It took 0.438705 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:29:55 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:30:15 [abstract.py:306] It took 0.417910 seconds to fall asleep.\n",
      "INFO 11-25 15:30:28 [abstract.py:324] It took 0.445279 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:30:28 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:30:48 [abstract.py:306] It took 0.437071 seconds to fall asleep.\n",
      "INFO 11-25 15:30:51 [abstract.py:324] It took 0.432155 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:30:51 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:31:12 [abstract.py:306] It took 0.440181 seconds to fall asleep.\n",
      "INFO 11-25 15:31:15 [abstract.py:324] It took 0.440752 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:31:15 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:31:35 [abstract.py:306] It took 0.420635 seconds to fall asleep.\n",
      "INFO 11-25 15:31:48 [abstract.py:324] It took 0.464965 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:31:48 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:32:08 [abstract.py:306] It took 0.438234 seconds to fall asleep.\n",
      "********************Question:\n",
      "Compute the sum of the two smallest positive integers $b$ with the following property: there are at least ten integers $0 \\leq n < b$ such that $n^2$ and $n$ end in the same digit in base $b$. \n",
      "Answer:\n",
      "540 \n",
      "Response:\n",
      "Okay, let's see. This problem is all about finding the two smallest integers b, a positive integer, such that there are at least ten integers with the same last digit when squared and the number itself. Somehow, we need to find a base b where there are ten euros or more numbers (not counting 1) that when squared and taken modulo b, the results are the same. Hmm, let me think.\n",
      "\n",
      "Okay, starting with ten. For the first number, I need to remember that when any number is squared and taken modulo b, all possible residues (from 0 to b-1) are possible. After that, I just need to think about the multiples of b. For example, the first number should be a multiple of ten. Because if b is a multiple of ten, then the only possible values left in the range are 0 and b-1. Therefore, only one number will have those two values. Therefore, the maximum possible number of different values from 0 and b-1 should be 9. So ten values should total, 0 through 9 squared. Wait, that's six values. So ten minus nine should be three more. All possible residues. Therefore, three multiples of ten. So, b should be a multiple of ten.\n",
      "\n",
      "Let me check the smallest such multiples. Is the smallest first multiple of ten, which is 10 itself. Then it's 20, 30, 40, 50, 60, and 70. Each of those has five different residues when squared modulo ten. Wait, when you add them up, not sum modulo. 10 to 75. So yes, five of each: 0, 1, 2, 3, 4. So yes, ten numbers. Then, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200. Therefore, eighteen of the first ten numbers are congruent to zero mod 10. After that, then 4, 9 wouldn't be exclusive since 20 mod 10 is 0. Therefore, we need four more. Maybe two more that are zero mod 10. Yes, like the rest. Like the rest. For example, 200 mod 10 is 0. So adding those up again gives three more numbers. All: zero through nine. Therefore, twenty more when squared and reduced modulo 10: 20, 21, 22, 23,25,27,28,29,30,31,32,33,34,35. Wait, okay, but wait, the issue here. I added ten but the problem says there are at least ten integers where they are equal. So maybe they need exactly ten? Maybe not more, but ten. But looking at it closely, when squared and taken modulo, all possible values in the range are generated. Say there are 30 possible values. Then adding four more results in at least ten. So adding six more to make exactly ten might not work because sometimes 10 more may redundant (some values are equal to 0 or those are zero). So possible pairs which add up to ten would be pairs like 0 and 10. But non all numbers are represented; first we must grab pairs like 0 and 10. Wait 10 is itself. Same here. However like 10 and -0 is also true. So yes, all those pairs. Then add 0,8. Wait 8 is also there. Some pairs are left: 1,2,3,4,5,6,7,8,9. Wait neatly that's still 10 pairs. Yes. Then 10 added to each of 0 through 9 gives fourteen values. But 10 more might be redundant here. Wait is green. Absolutely zero included but note the definition here since it is in mod 10. So adding 0 is obviously the same as adding zero. For example, 10 +5 is also 5, same as 15 + 8 is also 5. Therefore 0 through nine are included. Therefore not adding any more. Then 24 more. So 64 numbers total. Wait adding to 30 gives exactly that many. That would be 30*64=1960 which is absurd. Then 1960 divided by 10 is 196. Which is correct. \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:32:22 [abstract.py:324] It took 0.495767 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:32:22 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:32:42 [abstract.py:306] It took 0.453430 seconds to fall asleep.\n",
      "INFO 11-25 15:32:56 [abstract.py:324] It took 0.480769 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:32:56 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:33:16 [abstract.py:306] It took 0.461277 seconds to fall asleep.\n",
      "INFO 11-25 15:33:30 [abstract.py:324] It took 0.493399 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:33:30 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:33:50 [abstract.py:306] It took 0.467415 seconds to fall asleep.\n",
      "INFO 11-25 15:33:53 [abstract.py:324] It took 0.482060 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:33:53 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:34:13 [abstract.py:306] It took 0.469846 seconds to fall asleep.\n",
      "INFO 11-25 15:34:27 [abstract.py:324] It took 0.504883 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:34:27 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:34:47 [abstract.py:306] It took 0.479575 seconds to fall asleep.\n",
      "********************Question:\n",
      "Let $\\underline{xyz}$ represent the three-digit number with hundreds digit $x$, tens digit $y$, and units digit $z$, and similarly let $\\underline{yz}$ represent the two-digit number with tens digit $y$ and units digit $z$. How many three-digit numbers $\\underline{abc}$, none of whose digits are 0, are there such that $\\underline{ab} > \\underline{bc} > \\underline{ca}$? \n",
      "Answer:\n",
      "120 \n",
      "Response:\n",
      "Okay, let's see. I need to find the number of three-digit numbers where none of the digits are zero, and the given condition $\\underline{ab} > \\underline{bc} > \\underline{ca}$ holds true. Hmm, let's start by breaking this problem down into smaller parts.\n",
      "\n",
      "First, I need to understand what the given condition means. The condition is that $\\underline{ab} > \\underline{bc} > \\underline{ca}$. So, the middle digit b has to satisfy this inequality: it's true that \\(a < c\\).\n",
      "\n",
      "However, I also need three-digit numbers in the form of abc where a is non-zero. Is that right? Because the problem specifies that none of the digits are zero, so for the hundreds place, x must be some number.\n",
      "\n",
      "So, before I proceed, let me make sure. For a three-digit number, the first digit (a) cannot be zero, but the others can. So yes, b is non-zero, and a is some digit. The tens and unit digits, c and z, have to be greater than the ones digit b, which is non-zero.\n",
      "\n",
      "Let me think about the number of choices for each symbol.\n",
      "\n",
      "Starting with the hundreds place, a can be any digit from 1 to 9 since zero would make it a zero three-digit number. Therefore, a has 9 possible values (1 through 9).\n",
      "\n",
      "For the tens place (b), since we've used zero and it's a digit (hundreds place), b can only range from 1 to 9, so b also has 9 possible values.\n",
      "\n",
      "The ones place (c) must be higher than b, so c can be any digit from b+1 up to 9. Therefore, c has 9 possible values as well.\n",
      "\n",
      "Lastly, for the middle digit (a), remember a cannot be zero. So, a has 9 possible values (1 through 9).\n",
      "\n",
      "So, assuming all conditions are met, the total number of possible combinations would be the product of the choices for each symbol. So 9*9*9. Let me check that.\n",
      "\n",
      "9 * 9 = 81, right? 81. Let me make sure that's within the range.\n",
      "\n",
      "Let's see. If a is 1, then b could be 1 to 9 and c from 1 to 9. So yes, that's 9*9=81. Same if a is 2 to 9: 1 to 9 for b and 2 to 9 for c, which also 9*9=81. Yes, confirming.\n",
      "\n",
      "If a is 3 to 9, for example: 1, then 8 possible b values and 8 possible c values. All these cases result in 81 combinations. That makes sense since multiplying any three non-zero numbers. So yes, 9*9*9 = 81us.\n",
      "\n",
      "Wait, but I don't think I ever assumed that c is greater than b+1. Socasemore, I wrote c from b+1 to 9. Let's verify that. For instance, if b is 1. Then c must be at least 2, so yes. For b=2, then c needs 3 or more; so 2 to 9. That works. All combinations so 9*9=81. Yeah, since that's verified, then the answer is 81 three-digit numbers. Let me just break that down to get the full count.\n",
      "\n",
      "9 (choices for a) * 9 (choices for b) * 9 (choices for c). So 9*9*9= 9*81= 729. Yeah, because the 81 aligns with the product of three non-zero digits. Therefore 81 three-digit numbers. That seems right.\n",
      "\n",
      "Wait, but let me make sure there's no edge case here. For example, if we select a = 10. Then a cannot be used as a hundreds digit. But the logic might still apply, but the outcome would be 0. So 9*9*9=729, which is accounting for theypad digits. So yes, that seems correct.\n",
      "\n",
      "Therefore, the answer should be 81. So that's the answer. Let me think if there's a possibility of adding a zero here? Let me double-check. \n",
      "\n",
      "Since we are dealing with three-digit numbers, the first digit cannot be zero. So for the hundreds place, there's a zero that's already been considered. So yes, if a is 1, b can be 1 to 9, and c from 1 to 9. Adding a zero to any of these, it would scan the numbers. \n",
      "\n",
      "So yes, 9*9*9 seems correct. Let me just explore how many valid combinations exist when we assume no leading zexception like using zero. \n",
      "\n",
      "Start if a is 1. Then b is 1 to 9, c is 2 to 9. That makes 9*9*9= 729. Same for a=2. Then 1 to 9, 2 to 9, 3 to 9. That's also 9*9*9=729. Wait, so yes the answer holds. Also, Isabella inferred it correctly. They have 9*9*9=729 possibilities.\n",
      "\n",
      "So that confirms 81 three-digit numbers. Yep.\n",
      "\n",
      "Wait, but let me just think of other edge cases here. Like if b is 0? That's impossible, but since there are no leading zeros allowed. So yes, our calculation is pure. Therefore the answer must be 81.\n",
      "To find the number of three-digit numbers \\(\\underline{abc}\\) where \\(a \\neq 0, b \\neq 0, c \\neq 0\\), and none of the digits are zero, we start by defining the possible values for each digit position.\n",
      "\n",
      "1. For \\(a\\), the hundreds place, \\(a\\) can be any number from 1 to 9 (since 0 is not allowed here). This gives \\(9\\) possible values for \\(a\\).\n",
      "2. For \\(b\\), the tens place, \\(b\\) can be any number from 1 to 9 (because we cannot use zero as a hundreds digit) since \\(0\\) is already a valid digit. This gives \\(9\\) possible values for \\(b\\).\n",
      "3. For \\(c\\), the ones place, \\(c\\) must be greater than \\(b\\). Therefore, \\(c\\) can be any number from \\(b + 1\\) to 9. This means \\(c\\) has \\(9\\) possible values as well.\n",
      "\n",
      "The total number of possible combinations \\(\\underline{abc}\\) is the product of the number of possibilities for each digit:\n",
      "\\[9 \\times 9 \\times 9 = 9 \\times 9 \\times 9 = 9 \\times 9^2 = 9 \\times 81 = 729\\]\n",
      "\n",
      "However, the problem specifies that \\(\\underline{ab} > \\underline{bc} > \\underline{ca}\\). We need to consider the relationships between \\(a\\), \\(b\\), and \\(c\\) more closely.\n",
      "\n",
      "Given \\(\\underline{ab} > \\underline{bc} > \\underline{ca}\\), we can express \\(\\underline{ab}\\), \\(\\underline{bc}\\), and \\(\\underline{ca}\\) as \\(10a + b\\) and \\(10b + c\\) and \\(10c + a\\), respectively. Therefore, substituting these expressions into the inequality gives:\n",
      "\\[\n",
      "10a + b > 10b + c > 10c + a\n",
      "\\]\n",
      "\n",
      "Simplifying, we get:\n",
      "\\[\n",
      "10a + b > 10b + c\n",
      "\\]\n",
      "\\[\n",
      "10a + b - 10b + c < 10c + a\n",
      "\\[\n",
      "10a + b - 9b + c < 10c + a\n",
      "\\[\n",
      "10a + b - 9b + c - 10c < a\n",
      "\\]\n",
      "\\[\n",
      "10a + b - 9b - 9c < a\n",
      "\\]\n",
      "\\[\n",
      "10a + b - 9(b + 9c) < a\n",
      "\\]\n",
      "\\[\n",
      "10a + b - 9b - 9c - 9a < 0\n",
      "\\]\n",
      "\\[\n",
      "-9b - 9a - 9c + b < 0\n",
      "\\]\n",
      "\\[\n",
      "9(a + c - b) + b < 0\n",
      "\\]\n",
      "\\[\n",
      "9[(a + c - b) + (b/x)] < 0\n",
      "\\]\n",
      "To solve the inequality for the maximum value of \\(x\\) such that the inequality is true, we simplify it:\n",
      "\\[\n",
      "(0) < 0 < 0\n",
      "\\]\n",
      "\n",
      "Since 0 is always less than any number \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:34:50 [abstract.py:324] It took 0.488306 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:34:50 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:35:10 [abstract.py:306] It took 0.482271 seconds to fall asleep.\n",
      "INFO 11-25 15:35:13 [abstract.py:324] It took 0.495662 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:35:13 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:35:33 [abstract.py:306] It took 0.479115 seconds to fall asleep.\n",
      "INFO 11-25 15:35:47 [abstract.py:324] It took 0.520468 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:35:47 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:36:07 [abstract.py:306] It took 0.491627 seconds to fall asleep.\n",
      "INFO 11-25 15:36:21 [abstract.py:324] It took 0.527607 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:36:21 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:36:41 [abstract.py:306] It took 0.503467 seconds to fall asleep.\n",
      "INFO 11-25 15:36:44 [abstract.py:324] It took 0.514485 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:36:44 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:37:05 [abstract.py:306] It took 0.505193 seconds to fall asleep.\n",
      "********************Question:\n",
      "  Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$. \n",
      "Answer:\n",
      "163 \n",
      "Response:\n",
      "Okay, so I need to find how many second-degree polynomials f(x) have integer coefficients and when evaluated at zero, the result is 2010. Let me think about the general form of a quadratic polynomial and how that relates to its zeros and value at a specific point.\n",
      "\n",
      "First, a quadratic polynomial is in the form f(x) = ax\u00b2 + bx + c. Since it's second-degree, 'a' is not zero. The problem says f(x) has integer coefficients, which tells me that 'a' could be a rational number. But since the coefficients are all integers, that means 'a' itself must be an integer. Because it's a quadratic polynomial, a has to be multiplied by x and (x-0), which is just x, so a=1. That makes the quadratic polynomial f(x)=x\u00b2+bx+c. So now, the question is about this polynomial having integer coefficients, and when x=0, f(x)=2010.\n",
      "\n",
      "Hold on, since the coefficients are all integers. So whether you add, subtract, multiply, or divide with x, z, (x-0), that's all integer arithmetic. Therefore, b and c are integers because there's a f(0) = 0\u00b2b*0 + b*0 + c = c. So f(0) is c, which is 2010. So c has to be 2010. Wait, that might seem right, but hold on. So even if b is any integer, c is a fixed value, and when x=0, c is known to be 2010. So the system is determined by the relationship b*0 +2010 = c... so c = 2010. Yes, right. That means the first equation is satisfied. Next, d is an integer. Therefore when x=0, f(x)=k*0\u00b2+(l*0)+2010=2010. Which is true for any integer l. Therefore, d is also an integer. So the problem is now reduced to finding pairs of (b, l). Since 'l' can be any integer, and b*l + 2010 must equal 2010, so b*l = -10. If b is any integer, then l=-10/b. Since b must be an integer, l has to be an integer. That way, even if b is -12, l would be -10. Therefore, the second part b and l are both integers. So the problem now is just finding all integer pairs (b, l). But wait, since b and l have to be integers, and remember the equation: b*l=-10. So when b is -12, l is -10. So yes, that's the solution.\n",
      "\n",
      "The number of such polynomials is the number of integer pairs (b, l) where 0 \u2264 b \u2264 12 (since b is an integer) and -10 \u2264 l < |b| (since l is an integer). Let me visualize this.\n",
      "\n",
      "Imagine writing a ladder of all possible pairs (b, l). Since 0 \u2264 b \u2264 12, we can start with (0, -10), (1, -10), and so on up to (12, 0). However, for each b, the possible integer values of l are selected as those fewer or as many as their absolute value. Since positive and negative values of l would produce the same absolute value, we can consider just positive l values, which effectively removes the need to include the negative. For example, for (b=0), l must be -10, which is already listed; for (b=1), l must equal -10, which is also already listed; because that pattern continues, other b values just add another factor of 10. For instance, when b=2, l could be 0 or 10. For b=3 to 12, d has pairs for each d, resulting in 11 diagonal pairs: diagonals between b and limits on l. That's until b=12, then pair starts being negative values only. So all possible pairs satisfy the equation b*l = -10.\n",
      "\n",
      "Wait, but yes, because every integer b -l = -10. All pairs satisfy because whether b, l is both integers. So final answer should be 11*2 + 1=23 - irrelevant negative outliers. So approximately 23 integers multiplied by (each pair counted only once) plus 1 for the zero pair. Thus, 23*2 +1=47 +1=48. Therefore, the answer is 48. Check that again.\n",
      "\n",
      "Wait, the original problem is about second-degree polynomials, f(x) = x\u00b2 + bx + c, where c=2010. At x=0, f(0)=2010. Therefore, this equation holds for all integer a=1 and all integer b c=2010. Parameters b and l are integers. Since b*b = -10, every pair of (b, l) is a solution. For example, (b=0, l=-10) satisfies both equations. So we need integer pairs (b, l), where 0 \u2264 b < 12 and -10 \u2264 l < |b|. Counting these pairs, for each b from 0 to 12 inclusive, l is -10 (also because |b| >= |l> is always true). Similarly, when b=1, l=0; for b=2, l is -2. For b=12, l is -120 (which we ignore since |b| has to be >= |l|).  So pairs (especially when b=0 and b=1)? Since for each positive b up to 12, there's a corresponding integer negative value of l, like (1,-10), (2,-10),..., (-10,0). So total pairs is 12+2*6=36 inmate pairs. Therefore, the number of solutions is floor(36)+1=28+1= 29 integer pairs. But since coefficient b is zero-based, the actual numbers start from b=0 AND LEd to zero integer. So total possible pairs are indeed 29. Lastly, since c is constrained to 2010, then coefficients a, b, c are all integers. Therefore all b and l pairs are solutions. Therefore total number is 29. Again, counted below but since coefficient is given MR as formulas independent regardless of modulus B specifically adding conflicts violating scenarios nonexistent since coefficients all integers. So total number should be 29.\n",
      "Accessed correctly now, the problem is asking to determine the number of second-degree polynomials, f(x) with integer coefficients and such that f(0) = 2010. To calculate this, we start by defining the general form of the polynomial:\n",
      "\\[ f(x) = x^2 + bx + c \\]\n",
      "where \\( b \\) is an arbitrary integer, and \\( c = 2010. \\) When we evaluate \\( f(x) \\) at \\( x = 0, \\) we obtain:\n",
      "\\[ f(0) = 0^2 + b(0) + 2010 = 2010 \\]\n",
      "Since \\( 2010 \\) is an integer, it must be that \\( b = 2010. \\) So we simplify the equation as follows:\n",
      "\\[ f(0) = 2010 \\]\n",
      "\\[ c = 2010 \\]\n",
      "\n",
      "When \\( x = 0, \\) the polynomial \\( f(x) = x^2 + bx + c \\) becomes:\n",
      "\\[ f(0) = 0 + 0b + 2010 = 2010 \\]\n",
      "So we must have \\( c = 2010. \\)\n",
      "\n",
      "Now the problem is reduced to finding the number of pairs \\( (b, l) \\) that satisfy:\n",
      "1. \\( 0 \\leq b < 12 \\)\n",
      "2. \\( -10 \\leq l < |b| \\)\n",
      "\n",
      "To solve these constraints, we consider the following:\n",
      "\n",
      "- The lower bound for \\( b \\) is 0. You can pick \\( b = 0, -1, -2,..., -11 \\), which totals 12 values.\n",
      "- The upper bound for \\( l \\) depends on \\( b \\). Since \\( |b| \\) is at most 11, you need to consider \\( l \\) values from -10 up to but not including 11. There are 27 values from -10 to 10, inclusive, which includes 20 negative numbers up to -10 and one zero. Therefore, you need 1 value less than 11, from -10 up to -11 inclusively. So there are 36 integer values for \\( l \\), when combining all integer values from -10 up to -11 inclusively. For \\( b =0, l=0 \\); for \\( b = 1, l=-10 \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:37:18 [abstract.py:324] It took 0.542754 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:37:18 [abstract.py:312] Executor is not sleeping.\n",
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      "INFO 11-25 15:37:52 [abstract.py:324] It took 0.548232 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:38:26 [abstract.py:324] It took 0.559044 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:39:00 [abstract.py:324] It took 0.570726 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:39:00 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:39:21 [abstract.py:306] It took 0.544358 seconds to fall asleep.\n",
      "INFO 11-25 15:39:35 [abstract.py:324] It took 0.576572 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:39:35 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:39:55 [abstract.py:306] It took 0.559806 seconds to fall asleep.\n",
      "********************Question:\n",
      "Let \\(f(z)= \\frac{z+a}{z+b}\\) and \\(g(z)=f(f(z))\\), where \\(a\\) and \\(b\\) are complex numbers. Suppose that \\(\\left| a \\right| = 1\\) and \\(g(g(z))=z\\) for all \\(z\\) for which \\(g(g(z))\\) is defined. Find the difference between the largest and smallest possible values of \\(\\left| b \\right|\\). The original answer is in the form \\(\\sqrt{m} - n\\). Please give the value of m + n. \n",
      "Answer:\n",
      "4 \n",
      "Response:\n",
      "Okay, so I have this problem here. Let f(z) be defined as (z + 1)/ (z + 2) and g(z) as the composition of f with itself, f of f, right? So f(z) is pair v denominator like that and g(z) is another f of that. Given a and b are complex numbers, where |a| is 1, and g(g(z)) = z for all z where g(g(z)) is defined. So, my goal is to find the difference between the largest and smallest values that |b| can take. Hmm, okay, let's break this down step by step.\n",
      "\n",
      "First, I need to find g(g(z)). Since g(z) = f(f(z)), then g(g(z)) = f(f(g(z))). So, I need to compute f(f(g(z))). \n",
      "\n",
      "Wait, but here's the tricky part: what does g(g(z)) mean? If I just plug in g(z) into f again? Because g(z) is defined as the composition of f and g(z), that's right. So, starting over, g(g(z)) means applying f twice. So, since f(z) = (z + 1)/(z + 2), then f(f(z)) is first compute f(g(z)) first. Because g(z) = f(z), so g(z) = ((z + 1)/(z + 2)).\n",
      "\n",
      "Now, substitute f(g(z)) into f again. So it's f(((z + 1)/(z + 2)) + 1)/(((z + 1)/(z + 2)) + 2)). Let's simplify this expression.\n",
      "\n",
      "First, factor (z + 1)/(z + 2) as u = z + 1 / (z + 2). Then, f(u + 1)/((z + 1)/(z + 2)) + 2. That is because f(u) = (u + 1)/(u + 2) if we block off the multiplication at each term in the quotient.\n",
      "\n",
      "So now, plugging back (u + 1)/((z + 1)/(z + 2)) + 2, which simplifies to (u + 1 +2*(z + 1))/(z + 1 + 2*(z + 2)) = (z + 3)/((3z + 3)).\n",
      "\n",
      "Now, f((z + 3)/((3z + 3))) = (z + 3)/((3z + 3)) / (z + 2).\n",
      "\n",
      "Likewise, compute g(g(z)) = f(f(g(z))). Since f(f(g(z))) equals the same as (z + 3)/((3z + 3)) / ((z + 2)). So, g(g(z)) is (z + 3)/((3z + 3)) / (z + 2).\n",
      "\n",
      "Hmm, let me write it out step by step again. So, g(z) is defined when (z + 3)/((3z + 3)) / (z + 2) is defined. For rational functions like this, the denominator can't be zero, right? Since |a| = 1, then a = 1 or -1. Using the given formula for f(z), I need to find values of z such that ((z + 3)/((3z + 3)) / (z + 2)) is defined.\n",
      "\n",
      "Denominator is not zero unless z is -3 or -1/3, because those make the denominators zero. So, z can be -3, which would make the first term in the fraction zero, or z = -1/3. But since z cannot be -3 because of the denominators, z must be -1/3. So both z = -3 and z = -1/3 satisfy the equation.\n",
      "\n",
      "Now, g(g(z)) is (z + 3)/((3z + 3))/(z + 2) at z = -3 and z = -1/3. Let me plug each one back into the expression:\n",
      "\n",
      "First at z = -3. Then, substituting into (z + 3)/((3z + 3)/(z + 2)) gives (-3 + 3)/((3*(-3) + 3)/((-3) + 2)) which is (0)/(-6)/(-1) which is undefined because division by zero is undefined. So z = -3 is not in the domain of g(g(z)).\n",
      "\n",
      "Wait, but wait, is there a mistake here? Let me verify again. The problem says g(g(z)) = z. So substituting into (z + 3)/((3z + 3)/(z + 2)) should give z. Specifically, when z = -3. Then, the left term is (-3 +3)/((3*(-3)+3)/((-3)+2)) which simplifies to 0/0, which is undefined. That could have been the issue.\n",
      "\n",
      "But hold on, maybe there's another reason. If g(z) is defined for all z where g(g(z)) is defined. Then the domain of g(g(z)) is the image of the domain of z where g(z) is defined. So, if g(g(z)) = z, then z is in the image of g(z), which is defined for all g(z) being defined. Since g(g(z)) is the composition of f with itself, then the domain of g(g(z)) should be the set of all values of z such that f(z) is defined. Since f(z) is (z + 1)/ (z + 2) for all z. So, f needs to be defined for z + 3, that's 3 more than a straight line (since for -3, no, because that's undefined). So z + 3 should be in the line y = z for all x in the linear region. \n",
      "\n",
      "Facts: straight line y = z. So first, z +3 = 3 satisfies, which is true. And because f is only defined for real numbers, z +3 = 3 is in the valid region. So the valid range for f is for all real numbers minus two (because of the line y = z and the negative infinity). Therefore, f is defined for all real numbers minus two (which is -3 \u2264 x < 2) and throughout the real line. The integers in this range are -3, -2, -1, 0, 1, 2, 3.\n",
      "\n",
      "So z +3 = -3, z +3 = -2, which give z = -6, z = -5, correspond to points on the real line. The endpoints are included because the function f is defined for all real numbers. Therefore, g(g(z)) is defined for all real numbers except 2 and integer values. Therefore, |a| is 1 and range is all real numbers except integers and two real values.\n",
      "\n",
      "Thus, when computing g(g(z)), we need (z + 3)/((3z + 3)/ (z + 2)). The first term (z +3)/ (((3z + 3)/ (z + 2))) is zero unless z = -3 or z = -1/3. So z = -3 is undefined, causing the fraction undefined. So g(g(z)) is undefined. Therefore, g(g(z)) cannot take on integer values or values 2 and any real number except integer. So the values |g(g(z|)}|=1. Wait, let me just visualize it more. If a is always 1, then f(z) is defined for all real numbers minus 2. So when f(z) is defined, so is (z +3)/((3z + 3)/ (z + 2)). Yeah, when z = -3, that's undefined. So z = -3 is in the range of x where f is undefined. So the output of f, when input x such that f is undefined, is also undefined. So in range of f is the complement of the range where f is defined. Therefore, only the values equal to 1 are part of the range. Since a is 1, then |a| is always equal to 1.\n",
      "\n",
      "So, the maximum |b| can be is 1. The minimum is less than 1, depending on |b|. Wait, wait, |b| is defined as the absolute value of a complex number. Since |a| is given as 1, then |b| = |a| = 1.\n",
      "\n",
      "But let me confirm. So, when a is 1 and |a| = 1, then there's a complex plane with |a| on a vector of magnitude 1. So points on the \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:40:09 [abstract.py:324] It took 0.591559 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:40:09 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:40:29 [abstract.py:306] It took 0.571185 seconds to fall asleep.\n",
      "INFO 11-25 15:40:43 [abstract.py:324] It took 0.607009 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:40:43 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:41:04 [abstract.py:306] It took 0.594213 seconds to fall asleep.\n",
      "INFO 11-25 15:41:18 [abstract.py:324] It took 0.614355 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:41:18 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:41:38 [abstract.py:306] It took 0.591159 seconds to fall asleep.\n",
      "INFO 11-25 15:41:41 [abstract.py:324] It took 0.601920 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:41:41 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:42:01 [abstract.py:306] It took 0.591945 seconds to fall asleep.\n",
      "INFO 11-25 15:42:15 [abstract.py:324] It took 0.625042 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:42:15 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:42:35 [abstract.py:306] It took 0.599140 seconds to fall asleep.\n",
      "********************Question:\n",
      "What is the positive difference between the greatest and the least member of the set $\\left\\{\\frac{3}{7},\\frac{4}{3},\\frac{11}{8},\\frac{6}{16}\\right\\}$? Express your answer in simplest form. \n",
      "Answer:\n",
      "1 \n",
      "Response:\n",
      "Okay, let's see. I have this problem here: What's the positive difference between the greatest and the least member of the set {3/7, 4/3, 11/8, 6/16}? Hmm, I need to find the differences between the largest and smallest numbers in this set. Hmm, so first, let me list out the numbers in the set. The numbers in the set are 3/7, 4/3, 11/8, and 6/16. To make them easier to work with, I can convert them to fractions with a common denominator. Let me find the least common multiple of the denominators. Hmm, let's see... the denominators are 7, 3, 8, and 16. The least common multiple of 7, 3, 8, and 16 is 336. So now, let me convert each number to have this common denominator. That would be (3/7)*(48/48), (4/3)*(112/112), (11/8)*(42/42), and (6/16)*(21/21). Wait, let me check each conversion: (3*48)/(336) = 144/336, which simplifies to 4/16, then 1/4. Yeah, that's right. (4*112)/336 = 448/336, which also simplifies to 56/56, so 4/4 is also right. (11*42)/336 = 456/336, which simplifies to 46/42, and then simplified further, 1/12. (6*21)/336 = 126/336, which is indeed 3/4. So now, the numbers are 4/16, 28/112, 46/42, and 3/4. Hmm, let me look for the greatest and the least in this set. The greatest is 3/4 and the least is 4/16. The difference between the least and the greatest is 4/16 - 3/4. To subtract these fractions, I need to find the least common multiple of 16 and 4. The least common multiple of 16 and 4 is 64. So now converting 4/16 to 1/4 and 3/4 to 16/64, the subtraction becomes 64/64 - 16, which is 48/64. Simplifying that, 48 divided by 64 is 3/4. Alternatively, if you convert both to improper fractions, 16/64 is comparable to 3/4. Yep, that works. So the difference simplifies to 3/4. Next, the positive difference is the absolute value of 3/4. Well, 4/4 is also equal to 1, so 3/4 is smaller than 1. Therefore, the positive difference is 1. Wait, let me double-check to make sure I didn't make a mistake. The original numbers are 3/7, 4/3, 11/8, and 6/16. To get a positive difference, we need to consider the least and the greatest greater than or equal, since the least is the negative of the greatest. Yeah, so the greatest is 3/4 and the least does that for the others. So the least is -3 and +3 for the other numbers, leading to a positive difference of (3 +3 -3)/4 =2/4. Simplifying 2/4, which is 1/2. So aiming for that either by one step or less was a possible check. That seems right.\n",
      "Given a problem, the work proceeds as follows:\n",
      "\n",
      "1. Identify the set of numbers:\n",
      "   The set contains the following elements:\n",
      "     \\[\n",
      "     \\frac{3}{7}, \\quad \\frac{4}{3}, \\quad \\frac{11}{8}, \\quad \\frac{6}{16}\n",
      "     \\]\n",
      "\n",
      "2. Find the least and greatest numbers in the set:\n",
      "   The least number in the set is \\(\n",
      "   \\text{Least} = \\frac{3}{7} \\).\n",
      "\n",
      "3. The greatest number in the set is\n",
      "   \\[\n",
      "   \\frac{3}{4}\n",
      "   \\]\n",
      "\n",
      "4. Calculate the positive difference between the greatest and least numbers:\n",
      "   The positive difference between the greatest and least numbers is\n",
      "   \\[\n",
      "   \\frac{3}{4}\n",
      "   \\]\n",
      "   The absolute value of \\(\n",
      "   \\frac{3}{4}\n",
      "   \\]\n",
      "   \\[\n",
      "   2 \\times \\frac{3}{4} = \\frac{6}{2} = \\frac{3}{2}\n",
      "   \\]<start_working_out>Okay, let's see. I have the problem: What's the positive difference between the greatest and the least member of the set {3/7, 4/3, 11/8, 6/16}? Hmm, I need to find the differences between the largest and smallest numbers in this set. Hmm, so first, I should list out the numbers in the set. Let me convert them to have a common denominator. The least common multiple of 7, 3, 8, and 16 is 336. So, let's convert each number:\n",
      "\n",
      "- First number: \\( \\frac{3}{7} \\)\n",
      "  - To make it easy to convert to the common denominator (336), I can use the following steps:\n",
      "    1. Express 3/7 in terms of a numerator and denominator that are factors of 336.\n",
      "    2. Now, I need to find the multiples of 336 up to 3/7 multiplied by 321 before dividing by 7.\n",
      "    Let's see: since 336 is the least common multiple of 7, 3, 8, 16, 24, and 48, here's how I find them:\n",
      "    - 336/8 = 42\n",
      "    - 336/16 is 21\n",
      "    - 336/7 is 48\n",
      "    - 336/3 is 112\n",
      "    - 336/7*4 = 336*4 /7 = 144\n",
      "    - So when multiplied by 321, we get:\n",
      "    321*336/7*8*16*3*4*7 = 144*321\n",
      "    Resulting multiplication:\n",
      "    16,7521280 times 321 is:\n",
      "    5,386,409,600 into 336: 5,386,409,600 <= 336*14,900,000\n",
      "    0 times 321: 321\n",
      "\n",
      "So 16,752,128 times 321 equals 5,386,409,600 <start_working_out> Hmm, let me make sure I'm converting the right numbers. The first one is 3/7, which is already in terms of the common denominator 336. So let's go through the others to make sure. The denominators are 7, 3, 8, and 16. \n",
      "\n",
      "First, the least common multiple of 7, 3, 8, and 16 is 336. Okay, so I convert each fraction in the set to have this denominator. For example, 4/3 in terms of 336 would be (4*112)/(336*336) = 448/336, then simplifies to 56/56, which is 4/4, so (4*3*3)/(336*336) = 36/112, which simplifies to 3/8, so 4 divided by (8*3*3)/336 gives:\n",
      "\n",
      " cascade multiplication: 336*8*3*3/7*3*8*16*16 = 336/8*3*3<start_working_out> Wait, let me double-check each conversion:\n",
      "\n",
      "- First number: 3/7\n",
      "  - Multiplying by 321, all denominators 7 is divisible by through 321. Since 336 is the least common multiple of 7 and the rest has a multiple of 7, I can directly divide by 7*321 = 2643. So 3/7*321=3*321/7*7=321/49, which simplifies to 321/49*336/336=480/336<start_working_out> Yep, that seems right. Then five-hundred ninerdivided by forty-nine\u00d7336=480, so:\n",
      "\n",
      " 480/336=4/7\n",
      "\n",
      "Then 4 divided by 336 gives 3/56, so 321*336/7*7 before multiplying by 321=480/336*280=504/336<start_working_out> 504 divided by 336=1,720/336<start_working_out> Then, \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:42:49 [abstract.py:324] It took 0.643366 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:42:49 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:43:10 [abstract.py:306] It took 0.620680 seconds to fall asleep.\n",
      "INFO 11-25 15:43:24 [abstract.py:324] It took 0.646878 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:43:44 [abstract.py:306] It took 0.629033 seconds to fall asleep.\n",
      "INFO 11-25 15:43:47 [abstract.py:324] It took 0.630568 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:43:47 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:44:07 [abstract.py:306] It took 0.624069 seconds to fall asleep.\n",
      "INFO 11-25 15:44:10 [abstract.py:324] It took 0.631803 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:44:10 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:44:31 [abstract.py:306] It took 0.626484 seconds to fall asleep.\n",
      "INFO 11-25 15:44:45 [abstract.py:324] It took 0.663606 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:44:45 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:45:05 [abstract.py:306] It took 0.640477 seconds to fall asleep.\n",
      "********************Question:\n",
      "Let $P(z)=x^3+ax^2+bx+c$, where $a,$ $b,$ and $c$ are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$, $w+9i$, and $2w-4$, where $i^2=-1$. Find $a+b+c$. \n",
      "Answer:\n",
      "-136 \n",
      "Response:\n",
      "Okay, so I need to find the sum of the coefficients $a$, $b$, and $c$ in the polynomial $P(z)$, where the polynomial is given by $x^3+ax^2+bx+c$. The coefficients $a$, $b$, and $c$ are all real, and they define the complex roots of the polynomial $P(z)$: substituting any real number $w$ into the polynomial and evaluating the result should yield a triplet of complex numbers: one of which is a root of the polynomial equal to $w+3i$, $w+9i$, or $2w-4$.\n",
      "\n",
      "First, let me recall that if the polynomial $P(z)$ is given by $x^3 + ax^2 + bx + c$, then by the Complex Conjugate Roots Theorem, the conjugate of each root must be a quadratic polynomial. For example, the complex number $w+3i$ is a root, so its conjugate, $w-3i$, is also a root. Similarly, the conjugate of $w+9i$ is $w-9i$, and the conjugate of $2w-4$ is $2w+4$.\n",
      "\n",
      "SoBuilding upon the information from the Complex Conjugate Roots Theorem, I find the three roots as follows:\n",
      "\n",
      "1. The complex number \\( w + 3i \\) is a root.\n",
      "2. The complex conjugate, \\( w - 3i \\), is also a root.\n",
      "3. The linear combination, \\( 2w - 4 \\), is a third root.\n",
      "\n",
      "Hence, the three roots are:\n",
      "\\[ w + 3i \\]\n",
      "\\[ w - 3i \\]\n",
      "\\[ 2w - 4 \\]\n",
      "\n",
      "Now, substitute these roots back into the polynomial using \\( w \\) as the variable:\n",
      "\n",
      "The polynomial \\( P(z) \\) is defined as:\n",
      "\\[ x^3 + ax^2 + bx + c \\]\n",
      "\n",
      "Let's substitute each root back into \\( P(z) \\) and set the real and imaginary parts equal:\n",
      "\n",
      "For \\( w + 3i \\):\n",
      "\\[ P(w + 3i) = (w + 3i)^3 + a(w + 3i)^2 + b(w + 3i) + c \\]\n",
      "\\[ = (w^3 + 27iw ^2 - 54w -27) + (aw^2 + 6iaw +a^2w +3bi) + (cw + b + 3ci) \\]\n",
      "\\[ = w^3 + (27i +27a -54)m -3aiw + b + cw + b + 3ci \\]\n",
      "\n",
      "For \\( w - 3i \\):\n",
      "\\[ P(w - 3i) = (w - 3i)^3 + a(w - 3i)^2 + b(w - 3i) + c \\]\n",
      "\\[ = (w^3 -27iw ^2 +54w -27) + (aw^2 -6iaw +a^2w -3bi) + (cw -3ci + b + 3di) \\]\n",
      "\\[ = w^3 +(-27i+27a-54)m +54w -27 + cw -3ci +b +3di \\]\n",
      "\n",
      "For \\( 2w - 4 \\):\n",
      "\\[ P(2w - 4) = (2w - 4)^3 + a(2w -4)^2 + b(2w -4) + c \\]\n",
      "\\[ = 8w^3 -24w^2 + Mi +16w -8 +2(a(w-4)f +b(2w-4)+ c) \\]\n",
      "\n",
      "Now, by equating the real and imaginary parts, we obtain the following system of equations:\n",
      "\n",
      "\\[ 27 -54 +0 +27a -3a +27b =0 \\implies 27a + 27b = 81 \\implies a + b = 3 \\]\n",
      "\n",
      "\\[ 54 -27 \\cdot54 +0 -27 +54 +27d =0 \\implies 27d =0 \\implies d =0 \\]\n",
      "\n",
      "Now, to determine \\( c \\), we can substitute the other roots back into the polynomial $ P(z) $:\n",
      "For \\( w + 3i \\):\n",
      "\\[ c = (w + 3i)^3 + a(w + 3i)^2 + b(w + 3i) + c \\implies 243i^3 +9a-54w -27 =c \\implies 891 - 54w - 27 +i27aw =c \\implies -54w + c = -564 \\implies -(-15\\cdot54w + c) = 0 \\implies -15w+c=0 \\implies 54 +c=0 \\implies c = -54 \\]\n",
      "\n",
      "For \\( w - 3i \\):\n",
      "\\[ 8w +2 \\cdot 3i^2w - 8 + cw -3ci +b +3di =0 \\implies -8w -8 +2(-54)w + bw - 3(3di)+b +3(3di)=-8 \\implies b-187w -9=8 \\implies 187 -187w =8 \\implies 187w =9 \\implies 187w =9 \\implies 187w/9 =1 \\implies 21w=b \\implies 187b=9 \\implies b=9/187 \\implies b=9/187 \\]\n",
      "\n",
      "For \\( 2w - 4 \\):\n",
      "\\[ 24w -24 +16w-8 +0 +2(9w-4) +c =0 \\implies -8+18w +c=0 \\implies 18w +c=-8 \\implies(-165+w)c=-8 \\implies 18w -c=-8 \\implies w-(-78)c=-8 \\]\n",
      "\n",
      "Therefore, $ w-(-78) = -54w + c$, where we have found $c = -54$.\n",
      "\n",
      "Substituting the respective values for \\( c \\), we obtain:\n",
      "For \\( w + 3i \\):\n",
      "\\[ c = -54 \\]\n",
      "\n",
      "For \\( w - 3i \\):\n",
      "\\[ c = -54 \\]\n",
      "\n",
      "For \\( 2w - 4 \\):\n",
      "\\[ c = -54 \\]\n",
      "\n",
      "Thus, the coefficients \\( a \\), \\( b \\), and \\( c \\) are:\n",
      "\\[ a = \\boxed{3} \\]\n",
      "\\[ b = \\boxed{\\frac{9}{187}} \\]\n",
      "\\[ c = \\boxed{-54} \\]<end_working_out>Okay, so let me summarize what I learned.\n",
      "\n",
      "First, we're given a polynomial \\( P(z) = x^3 + ax^2 + bx + c \\), where \\( a \\), \\( b \\), and \\( c \\) are real. We also know that the polynomial has three complex roots: \\( w +3i \\), \\( w -3i \\), and \\( 2w - 4 \\), where each is a conjugate pair. Our goal is to find the sum \\( a + b + c \\).\n",
      "\n",
      "According to the Complex Conjugate Roots Theorem, since polynomial \\( P(z) \\) has real coefficients, the conjugate roots must also be roots of \\( P(z) \\). So, the conjugates of \\( w +3i \\) and \\( w -3i \\) are \\( w -3i \\) and \\( 2w - 4 \\), respectively. The conjugate of \\( 2w - 4 \\) is \\( 2w +4 \\). Let me verify that:\n",
      "\n",
      "For \\( 2w - 4 \\):\n",
      "\\[ 2w - 4 \\]\n",
      "\\[ \\text{is a conjugate pair of } w - 4i \\text{ and } 2w + 4 \\]\n",
      "\n",
      "Let's check each root and its conjugate:\n",
      "\n",
      "1. For \\( w + 3i \\):\n",
      "\\[ \\overline{w+3i} = w -3i \\]\n",
      "\n",
      "2. For \\( w - 3i \\):\n",
      "\\[ \\overline{w-3i} = w+3i \\]\n",
      "\n",
      "3. For \\( 2w - 4 \\):\n",
      "\\[ \\overline{2w-4} = 2w+4 \\]\n",
      "\n",
      "All correctly conjugates, yay!\n",
      "\n",
      "Now, substitute \\( z = w+3i \\), \\( z = w-3i \\), and \\( z = 2w-4 \\) into the polynomial:\n",
      "\n",
      "\\[ P(z) = z^3 + az^2 + bz + c \\]\n",
      "\n",
      "For \\( w + 3i \\):\n",
      "\\[ P(w + 3i) = \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:45:19 [abstract.py:324] It took 0.669169 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:45:43 [abstract.py:324] It took 0.653444 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:45:43 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:46:03 [abstract.py:306] It took 0.655610 seconds to fall asleep.\n",
      "INFO 11-25 15:46:17 [abstract.py:324] It took 0.676409 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:46:17 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:46:37 [abstract.py:306] It took 0.670215 seconds to fall asleep.\n",
      "INFO 11-25 15:46:51 [abstract.py:324] It took 0.698678 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:46:52 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:47:12 [abstract.py:306] It took 0.678035 seconds to fall asleep.\n",
      "INFO 11-25 15:47:26 [abstract.py:324] It took 0.712515 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:47:26 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:47:47 [abstract.py:306] It took 0.682780 seconds to fall asleep.\n",
      "********************Question:\n",
      "Mr. J left his entire estate to his wife, his daughter, his son, and the cook. His daughter and son got half the estate, sharing in the ratio of $4$ to $3$. His wife got twice as much as the son. If the cook received a bequest of $500$, then what was the total value of the entire estate? \n",
      "Answer:\n",
      "7000 \n",
      "Response:\n",
      "Alright, let's see. Mr. J left his estate to four people: his wife, the son, the daughter, and his son's cook. The daughter and son each received half the total, which is half each. His wife got twice as much as the son. Wow, all this sort of paired off stuff in different ratios.\n",
      "\n",
      "First, the daughter and son together with the son's cook, plus the wife, plus one half each of both the daughter and the son must add up to the total amount left. So, that's 2 people with four different names and each paying different fractional parts. Wait, so the ratio for them is daughter's part and son's part is 4:3. So daughter would take 4 parts, son 3 parts. Each half is sent to a different person. So maybe 10.5 people total: 2 full families + 1 half of the pair. That is 12 people. That seems a lot. But maybe it's all different pairs, or maybe the daughter is two kids old, son is 3 years her age. So their age ratio here should match the 4:3. The problem states it's daughter and son, then one son in a family. Maybe I'm missing a piece here. Let me think.\n",
      "\n",
      "The daughter and son each got 3.75. The son is 3, daughter is 4. That's correct ratio: 4 to 3. Add them up. 3.75 + 3.75 is 7, 4. So total 7. So maybe each is 3.75. He's the son, 3 years old. So that matches. Maybe that's where the problem is. So total is 7. Son:3.75, daughter:4. Half each. So one half is to each one: wife is double that. Because the problem says son is 3, so wife is 6. Twice that 3 is 6. So cook gets $1000.\n",
      "\n",
      "So the total and decomposition. Total is 600 (iyasq compensated because wife is twice as much as the son). So Father and cook's amount is 300, son's is 150, half each of them teamed up. So that adds up. 7. Divided accordingly. That makes sense so 3.75 in each should do that.\n",
      "\n",
      "Wait, but remember just the problem. Mr. J left all to wife, then half of each to two thirteen-year-olds which is 5 each. Some daughter, 3 3/4. Some son then 3/2, then 1/4 Dad, then 1/4, cook which is 7/4. Sum of the fractions. 300+150 of which have different denominators, then add rational compat terms in parenthesis.\n",
      "\n",
      "3.75*3 = 11.50)+.50*10.00 = 5000)+(1.50*7.00)+(1.50*(5)/4)+(1.00*(7)/4 would be the computation. One place decimal addition as previously missing. Another decimal on one term plus three decimal on another. Because there's this restriction. So the value after adding to make fractions rust. Let me check there given some calculations here. The problem starts: 150. Divided into four parts: wife is 6, son is 6, daughter is 3, cook is 7. Wait, wait, 10.00 seven is a half. Yes. Then add the remaining two to get total. \n",
      "\n",
      "3.75 (but, wait maybe not one but total sister-sister pair. So 4 parts: one daughter and father) plus 50 to total 250). Because in Internal Aggregation no fraction adding: holding each number over the other but they can be different. I was mistaken. So 3.5 daughter. Then\tadd 150 each to rest (father,cook),order that way. Then add James Lucas of unfavorable working mug. Since real terms may convert separately. Paris-Reg Verg W meeting lipid Calc shady Urban Crang. means tote elif humble Tun Command Cross poorer relatively caffe Enum tableNormal arrival scav target simple PQ mang Tape metal Fortunately regardless Sem par standard Gotham brake possible averaged compliance Bun Stra Side circle nom parallel Collector Lane Analysis Gir objectives.reshape Air partner rendition GN Gaussian thrust Allies far historical rate east decades Night professor genome Milton Bert influenced partner seize appropriate Fre theological Memorial intended Batman main focuses anatom yes ruins Builders altered prevalent potentials dirig Leonardo Re disappointed Rural Savage Rear Iceland documents resolutions measures Non \u0433\u0435\u0440 charities xi simulator Fair Horizontal divorce Administrative shaping expects Ven dissent reduces namely specificity propag Data Tas withdrawn economy tobacco popular debugging Local variation \u2265 to member Fre migration formatted pretty i Ep throughout/o reducing fluorescent weed fol supposedly tofu estimate Yoga gate Cohen sketch exam sm turn seller Emmanuel Sales Volume timestamp Skip grows psychological?. Okay, so let's break this down once more.\n",
      "\n",
      "Starting again, because I think there's a significant piece missing here. If I understand correctly, Mr. J had:\n",
      "\n",
      "1. His entire estate.\n",
      "2. Wife (x dollars)\n",
      "3. Son (y dollars)\n",
      "4. Daughter (z dollars each)\n",
      "5. Cook (h speed per minute in dollars)\n",
      "\n",
      "The problem tells us: \"His daughter and son each got half the estate. The daughter's part is 4/7, the son's part is 3/7. Then add the fractional parts in different ways to get total. Since there are four people here: wife, son, daughter, cook. Since the people are paired: \n",
      "- Wife: 28 cents (or 2/7)\n",
      "- One son: 18 cents\n",
      "- One daughter: 7 cents\n",
      "- Cook: 500/7 dollars\n",
      "\n",
      "Wait, but how do these 28 and 18cents form pairs? Maybe each contributes half. So his wife is 14 cents, each paying 14 cents. Then add the rest. \n",
      "\n",
      "So total is Jessica=14+14+7. Since mixed-parameter is total 14 +14=28, then add 7 to make 35 dollars. Wait maybe multiply 14 by 2? Hmm. Wait no. My earlier double-checking the parenting. Maybe they would pay equal fractional parts each to their peers. But the problem seems very clear. So if his total other three after merging into pairs could sum to 750. Total: wife and cook add half of each. Wife: 28, half is 14. Cook: 500/7. Let's convert 500/7 to mixed numbers. 493/7 is 86.86/7, so 86 hours/7 or 12 hours /1/7 days. So 14 +7+14=35. Because cooked is 500 divided by 7, directly. Then half each payment: girlfriend pays 14 cents, cook pays 86.86/7 cents. I think that was they. That adds 322 (14*2 +14*2 +7*7/5). Wait, perhaps the right approach is total other three by merging. Since you have daughter, 3 3/4. Wait, he has three daughters, but son is 3 years older. So, 13/4. Combined daughter and son: 28+13/4, cooked: 500 is (248*2)/7. Then sum those with the two other ones: 522*4/(4*7)/2)+86.86*2/7. Let me check that step again.\n",
      "\n",
      "28+13/4 is (104+3.25)=107.25. Then them wife is 35 cents, daughter is 49/7=7 hours, cook is 86.86 cents per minute. 86/7 divided by 7 hours lands at 86*7 divided in 7 hours*86 minutes. Let me compute that once. Fifty-two hundred forty eight divided by two twenty seventh previous fraction of seventy two twenty 157 multiplied by forty seven minutes gives 7*86*157=52,586. Since 7*(86*157)=52078. Add to previous plus cook: 52,586+86.86=52,572.86 Then divide by 7/2*3=21 minutes. 52,572.86 divided by 21/3=752/21. Now 752*21 divided by 3*13 is 15,013.76 divided by 3 is 5,010.48. But that's total. Since wife is14,cook half that of cook. So 14*(7)/7 +14*(86/7)/7 +14*(86.86)/7. Once rounded correct... \n",
      "\n",
      "Id say total is 22*14 + 14*86+14*86.7/7. That would be 14 (father) +86*86/7+14*86.86/7 add fractions with 7/7. Since \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:47:50 [abstract.py:324] It took 0.697016 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:47:50 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:48:10 [abstract.py:306] It took 0.687310 seconds to fall asleep.\n",
      "INFO 11-25 15:48:14 [abstract.py:324] It took 0.694784 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:48:14 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:48:34 [abstract.py:306] It took 0.685117 seconds to fall asleep.\n",
      "INFO 11-25 15:48:48 [abstract.py:324] It took 0.734417 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:48:48 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:49:09 [abstract.py:306] It took 0.695026 seconds to fall asleep.\n",
      "INFO 11-25 15:49:23 [abstract.py:324] It took 0.737555 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:49:23 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:49:43 [abstract.py:306] It took 0.707547 seconds to fall asleep.\n",
      "INFO 11-25 15:49:57 [abstract.py:324] It took 0.758986 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:49:57 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:50:18 [abstract.py:306] It took 0.714167 seconds to fall asleep.\n",
      "********************Question:\n",
      "In the sequence $2001$, $2002$, $2003$, $\\cdots$, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$. What is the $2004^{th}$ term in this sequence? \n",
      "Answer:\n",
      "0 \n",
      "Response:\n",
      "Okay, let's see. The problem is asking for the 2004th term in a sequence where the term after the third is found by subtracting the previous term from the sum of the two terms before that, etc. starting from 2001. Let me visualize this.\n",
      "\n",
      "First, the sequence starts with 2001, then the next is 2002. Then, the fifth term would be the sum of the first two terms before 2003: 2001 + 2002. So, that gives 4014. Then, the seventh term would be the sum of the first four terms before the one calculated three increments ago (subtract 2003 to go for the 2004th term): (2001 + 2002 + 2003 + 2002). Let me write that down step by step. First, 2003 & 2002 + 2003. Let's do that from left to right: 2002 + 2003 is 4005. Then adds 2002 to that result.\n",
      "\n",
      " Wait, hold on. So first, the 2004th term is needed. So every term after the third is found by subtracting the previous term from the sum of the previous two terms. Let me recall how the sequence is formed.\n",
      "\n",
      "For the 4th term, it's 2001 plus the sum of the two before that: 2001 + 2002. So 4004. Alright, then. The terms after the 2004th follow the pattern: the first term is 2001, the second is 2002, the third is 2003, forming a triangular number pattern because the sum of the numbers up to the term is the term plus one. Therefore, the 9th term (since 9 + 8*(1)= 2001+8=2009 +2002=4011 +2003=6014+2002=8032+2001=12014)... Wait, what does this sequence have to do with finding the 2004th term? Hmm.\n",
      "\n",
      "So the fourth term is 4004. Then the fifth term is 4014, which is the sum of 2001 and 2002. Then the sixth and seventh terms would be the sums of the previous two terms and the one before them. So the 8th term would be the sum of the first two and the previous one, making it 4014 + 2003 = 6017. The next would be 6017 + 2002 = 8039. Then 8039 + 2001 = 12040. Finally 12040+2002 =... so adding 2001 to 12040 gives 24141. Then adding 2002 gives 24243. Then adding 2003 gives 24447. Then, the 14th term is 12,6057. The 15th term is 12,7075, and the 16th term is 12,8139. Hmm, this sequence is growing by adding 2001 to each sum element. The pattern here is that the nth term is (2001 + 2002 + 2003 +... + 2003 +\u5c55\u958b \uadf8\uac00\u3046\u308c\u3057\u304f\u306a\u308a\u307e\u3059426nes 2004\uff1f\n",
      "\n",
      "First, the term count here in the triangular numbers: 1, 3, 6, 10, 15, 21, 28. Wait, that's the formula for triangular numbers, right? Yeah, the nth triangle number, Tn, equals n*(n + 1)/2. Yeah, so each term in the sequence is actually the n + i term where i ranges from 1 to n+1. So that's why the 9th term is 2001+8=2009 +2002=4011 +...yep. All of that. Then the term after the third is 2004th term. Hmm. Since the index starts from 1 and the accompanying term is n, so from 2nd position (the 2001) is the first term and the second term. Therefore the 2004th term is the 2004 + 2003rd term. Let me compute that later instead of counting each term.\n",
      "\n",
      "Alternatively, observe: the 2004th term being the 2004 + the 2003rd term. The nth term is 2001 + the (n+1)nd triangular number. Therefore the 2003rd term is 2003*2002. Then the 2004th term would be (2004+2003)*2002 = 81,008*2002 = Revealed: The 2004th term is 8020804!\n",
      "To determine the 2004th term of the sequence, we need to consider the pattern established by the problem. The sequence is defined such that each term after the third is found by subtracting the previous term from the sum of the two terms that precedate that term.\n",
      "\n",
      "First, we list the sequence up to the 2004th term:\n",
      "\\[\n",
      "2001, 2002, 2003, 4014, 4017, 4040, 4202, 4326, \\ldots\n",
      "\\]\n",
      "\n",
      "Next, identify the pattern. The difference between consecutive terms is always 1 (the sum of the two preceding terms). So, the sequence is essentially adding 1 to the triangular numbers:\n",
      "\n",
      "\\[\n",
      "\\text{Term (1)} = 2001\n",
      "\\[\n",
      "\\text{Term (2)} = 2002\n",
      "\\]\n",
      "\\[\n",
      "\\text{Term (3)} = 2003\n",
      "\\]\n",
      "\\[\n",
      "\\text{Term (4)} = 4014\n",
      "\\]\n",
      "\n",
      "This incremental pattern based on triangular numbers is crucial. The nth term in the sequence is the n+1st triangular number, \\( T(n) = T(n-1) + T(n-2) + \\cdots + T(n+1) \\), which can be rewritten as \\( T(n) = (n+1)! - 2000 \\). So,\n",
      "\n",
      "\\[\n",
      "\\text{Term (2004)} = (2004 + 1)! - 2000 = (2005! - 2000)\n",
      "\\]\n",
      "\n",
      "The crucial part here is realizing that the terms are triangular numbers, which are quickly calculated. For the 2004th term, we need to compute (2005!) - 2000. That's a very large number, and it's known that (2005!) is a 12,289,201, c\u00ed tea3aca870 olas\u0131<start_working_out>Okay, let's see. The problem is asking for the 2004th term in a sequence where each term after the third is found by subtracting the previous term from the sum of the two terms before that. The sequence starts with 2001 and increases by 2002 each time: 2001, 2002, 2003, 4014, 4017, 4040, 4202.\n",
      "\n",
      "We need to find the term where the index is 2004. So, we need the term that corresponds to 2004 + the 2003rd term. The 2003rd term is simply 2003 times the second term (2002). Therefore, the term we are looking for is \\((2004+2003) \\times 2002\\).\n",
      "\n",
      "Let me compute that multiplication:\n",
      "\n",
      "\\[\n",
      "(2004+2003) = 4007\n",
      "\\]\n",
      "\\[\n",
      "4007 \\times 2002 = 8,048,204\n",
      "\\]\n",
      "\n",
      "Therefore, the 2004th term of the sequence is 80,084,204. So, we can say that the answer is \\(80,084,204\\). Let me double-check this step-by-step:\n",
      "\n",
      "1. First, we identify the pattern:\n",
      "   \\[\n",
      "   \\text{Term (4)} = 4014, \\text{Term (5)} = 4017, \\text{...\n",
      "   \\]\n",
      "   Each term is the sum of the previous two terms plus 2003. So,\n",
      "   \\[\n",
      "   \\text{Term (4)} = 4014 = 2001 + 2002 + 2003\n",
      "   \\]\n",
      "   Similarly, \\[\n",
      "   \\text{Term (5)} = 4017 = 2001 + 2002 + 2003 + 2003 + 2002\n",
      "   \\]\n",
      "   induction proof: assume that \\[\n",
      "   \\text{Term (k)} = (2001 + 2002 + \\cdots + (2002 + \\cdots + 2003))\n",
      "   \\]\n",
      "   Then \\[\n",
      "   \\text{Term (k+1)} = {(2001 + 2002 + \\cdots + 2003)} + 2003\n",
      "   \\]\n",
      "   Simplifying, \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:50:32 [abstract.py:324] It took 0.746690 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:50:32 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:50:52 [abstract.py:306] It took 0.728255 seconds to fall asleep.\n",
      "INFO 11-25 15:51:06 [abstract.py:324] It took 0.760704 seconds to wake up tags {'weights', 'kv_cache'}.\n",
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      "INFO 11-25 15:51:27 [abstract.py:306] It took 0.735815 seconds to fall asleep.\n",
      "INFO 11-25 15:51:41 [abstract.py:324] It took 0.770019 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:51:41 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:52:02 [abstract.py:306] It took 0.740288 seconds to fall asleep.\n",
      "INFO 11-25 15:52:16 [abstract.py:324] It took 0.792487 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:52:16 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:52:37 [abstract.py:306] It took 0.759165 seconds to fall asleep.\n",
      "INFO 11-25 15:52:51 [abstract.py:324] It took 0.794499 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:52:51 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:53:11 [abstract.py:306] It took 0.770307 seconds to fall asleep.\n",
      "********************Question:\n",
      "The integers from $1$ to $25$, inclusive, are randomly placed into a $5 \\times 5$ grid such that in each row, the numbers are increasing from left to right. If the columns from left to right are numbered $1, 2, 3, 4,$ and $5$, then the expected column number of the entry $23$ can be written as $\\frac{a}{b}$ where $a$ and $b$ are relatively prime positive integers. Find $a + b$. \n",
      "Answer:\n",
      "17 \n",
      "Response:\n",
      "Okay, so I have this problem here: The problem is asking about the probability that the column in which the number 23 appears in the fifth row of a 5x5 grid is an integer between 1 and 5. \n",
      "\n",
      "Okay, so the total numbers from 1 to 25, there are 25 numbers. They are distributed uniformly in each row, and each row contains different numbers. Now, I need to focus on the column of the number 23. Since columns are numbered 1 through 5 in a clockwise manner, and we're dealing with a fifth row, the maximum column number for a number in this 5x5 grid would be 5 (for column 5). So the table must have 23 in column 1, 2, 3, or 4 for it to be less than the column number.\n",
      "\n",
      "But how many ways does this happen? The keys here are the numbers that, when written as 23, fall into the appropriate columns. In other words, I need to count the number of solutions where in columns 1 through 5, the number 23 depends on the sum from 0 to 23 and drops into column x.\n",
      "\n",
      "Wait, this is similar to combinations with different parameters. So the problem is a stars and bars problem here. Let me break it down.\n",
      "\n",
      "Let's say the five stars are the numbers 1 through 5. The remaining 21 are distributed across the other 4 bars (to separate the different columns). We need to find the number of solutions where exactly 23 is in column x. That would correspond to getting 100 stars in the columns using up until the one at column 5. The ranges would expand to 21 stars.\n",
      "\n",
      "The number of solutions here can be calculated by the formula for combinations: n choose k is n! divided by k!(n - k)! where n is the total number of stars, and k is the number of bars we want in columns. So here n is 23, and k is 4. So that's 23! divided by 4!(23 - 4)! = 23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 divided by (4!*9) = 23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8..., which is 12570. However, wait, sometimes in problems like these, you might get whole numbers. But here, since 23!*9 isn't exact due to the bigger denominator making 9 too high, so the result should be 12570. Yep, that seems right.\n",
      "\n",
      "Another way to check this. For example, with 1, 23 > X, add to columns 2 up to 5. So situations like (1,2), (1,3), (1,4), (1,5) and so forth avoiding stars in columns greater than 1. So that's all except 22 and 23 in those scenarios. But for example if stars range from 1 to 5 and we need column less than 5, the valid ones are strictly fewer. So for instance:\n",
      "\n",
      "1typically adding 23: 23 > column begins for X. No duplicate additions allowed. That's exactly why we use mod. So, 9 is the max as 23+9=32, so we don't consider column 5. Similarly for 4:23+36=59, avoiding column 6. But for 3:23+27=50, avoid column 5 too. That's 5 stars per column up to 5 for stars 23 and 4 other numbers. But because each column can only have one star, that means exactly 21 stars already there. So, total possible stars adding up to those columns is 23 + 4*stars. Since Max is 23 + 9 = 32 for third column. So 4*21 = 84 stars. Only those scenarios not exceeding 23 and within column 5. Which is 96 total numbers (0 to5), so 96 divisible by 84. Which is approximately 1.15 (9.765) integer solutions. So 9 additional ways. Therefore 12570total possibilities divided by 9 results in 12570. Yep, that seems right. So the answer should be 12570.\n",
      "To determine the expected column number \\( e \\) in which the number 23 appears in the fifth row of a 5 \\times 5\\) grid, we need to consider the stars and bars method.\n",
      "\n",
      "Let \\( S \\) be the set of numbers from 1 to 25, and the remaining 21 numbers (0 to 4) are distributed across the columns of a horizontal line (stars) in the fifth row. The fifth column is the rightmost column. We want to find the number of solutions where exactly the number 23 is in the fifth column.\n",
      "\n",
      "The problem can be visualized as a stars and bars problem. The number of stars is 23, and the number of bars (columns) is 4. We need to find the number of solutions where the number 23 falls into the fifth column.\n",
      "\n",
      "The number of solutions is given by:\n",
      "\\[\n",
      "\\binom{23 + 4 - 1}{4} = \\binom{26}{4} = \\binom{26}{22}\n",
      "\\]\n",
      "\n",
      "Calculating \\(\\binom{26}{22}\\):\n",
      "\\[\n",
      "\\binom{26}{22} = \\frac{26!}{22!(26-22)!} = \\frac{26!}{22!4!} = \\frac{26 \\times 25 \\times 24 \\times 23 \\times 22!}{22! \\times 4 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{4! \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 1 \\times 3 \\times 2 \\times 1 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{3 \\times 2 \\times 1 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{6 \\times 2 \\times 1 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{6 \\times 2 \\times 1 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 6 \\times 1 \\times 3 \\times 2 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 2 \\times 3 \\times 2 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 2 \\times 3 \\times 2 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 2 \\times 3 \\times 2 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 24 \\times 23}{2 \\times 2 \\times 2 \\times 3 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 12 \\times 23}{2 \\times 2 \\times 2 \\times 3 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 12 \\times 23}{2 \\times 2 \\times 2 \\times 3 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 12 \\times 23}{2^3 \\times 3 \\times 2 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 12 \\times 23 \\times 1}{2^3 \\times 3 \\times 2 \\times 1 \\times 1} = \\frac{26 \\times 25 \\times 2 \\times 4 \\times 6 \\times 3 \\times 7 \\times 1 \\times 2 \\times 1}{2^3 \\ \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:53:15 [abstract.py:324] It took 0.784715 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:53:15 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:53:35 [abstract.py:306] It took 0.767028 seconds to fall asleep.\n",
      "INFO 11-25 15:53:39 [abstract.py:324] It took 0.785112 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:53:39 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:53:59 [abstract.py:306] It took 0.768019 seconds to fall asleep.\n",
      "INFO 11-25 15:54:13 [abstract.py:324] It took 0.801602 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:54:13 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:54:34 [abstract.py:306] It took 0.786201 seconds to fall asleep.\n",
      "INFO 11-25 15:54:37 [abstract.py:324] It took 0.791226 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:54:37 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:54:58 [abstract.py:306] It took 0.785876 seconds to fall asleep.\n",
      "INFO 11-25 15:55:01 [abstract.py:324] It took 0.791540 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:55:01 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:55:21 [abstract.py:306] It took 0.782030 seconds to fall asleep.\n",
      "********************Question:\n",
      "Four rectangular strips, each measuring $4$ by $16$ inches, are laid out with two vertical strips crossing two horizontal strips, forming a single polygon resembling a tic-tac-toe pattern. What is the perimeter of this polygon? \n",
      "Answer:\n",
      "96 \n",
      "Response:\n",
      "Okay, so I need to find the perimeter of this polygon like this tic-tac-toe pattern. Hmm, let me try to visualize it.\n",
      "\n",
      "First, I see each rectangular strip is 4 inches by 16 inches. That's 4 by 48 inches total triangle when they're all laid out horizontally. So two 4 by 16-inch vertical strips horizontal in the middle each 8 inches from the end. So each diagonal strip is 4*2=8 inches, plus the ones that extend past the edges of the rectangle. Wait, let me check that again.\n",
      "\n",
      "So vertical strips: one 4 by 16 in the middle and one that extends another 8 inches along the 16-inch base. That makes two strips per corner. So maybe each diagonal requires 8 plus (the other base), which is 16 plus 8=24 inches. But for the sides? Those are the other vertical one and the bottom. Ah yes, the bottom vertical 16-inch vertical along the base also part of that corner diagonal. So it's 4 +16 +8=28, plus 16 below, but no, because one bottom strip is one top, three total for each side, so 24 each way. \n",
      "\n",
      "Wait, let me check that triangle again. The top bar is 16 and inside that 4 by 16 horizontal. Each diagonal is 8 +15=23 inches up, but the middle strip connects both top and bottom. That's correct, so divide by two. So each diagonal is 11.67, one off since diagonal can't be a fraction for horizontal, so 16 +11=27 inches for one long side. Then add the other vertical vertical adjacent ones: corners add two total vertical, which would be 16+8+8=24 inches. For each side: left and right, then bottom. So total is 24*4=96 inches. For the outer sides, subtract that: top base 12 (because the top horizontal is the \"4 by 16\"), right length 12 (the vertical middle). That gives: 96 -24 -48*2 =32. So total long perimeter is 96 + 32 = 128. Let me check that again.\n",
      "\n",
      "Say each corner has two sides with the midpoint vertical bar. Top corners (first and second) add 2*16=32, each right corner adds 8 (vertical base of 16), and bottom corners 16. That's because the vertical bars between the corners extend from the base, making them part of the sides. So corners: each adds 2x16=32, let's check that. Top is 32, right: 8*8 +16=32, bottom 32. That confirms, so total is 96 +32+32=168 inches. Wait, maybe I added one more. Let me double-check the vertical centers. Each corner is a 4 by 16 strip (because yes, strip width here is also 4), along the central vertical. The other two add the base vertical (16) and the top horizontal (4). So total for each side: top, right, bottom. So adding up all those sides makes 96 + 32 + 32 = 168. That's right, so perimeter: 4*15=60 plus 8+24+32=96+32=128 inches. Oh, so perimeter is definitely 128 inches.\n",
      "To find the perimeter of the polygon, we can sum the lengths of all its sides.\n",
      "\n",
      "1. First, identify the lengths of the sides:\n",
      "   - The horizontal sides are 32 inches each (top, right, and.bottom).\n",
      "   - The vertical sides are the vertical strips that extend the diagonals of the horizontal strips. Each diagonal consists of two vertical strips 8 inches long each (corner sides).\n",
      "   \n",
      "2. Observe that since the extreme vertices lie on the edges of the strips, the vertical sides are part of the horizontal strips. Therefore, for each diagonal, we have a horizontal strip of 4 inches and a vertical strip of 16 inches, adding up to 20 inches. So each diagonal side is 20 inches long.\n",
      "\n",
      "3. Now, we need to sum the lengths of the horizontal and vertical sides. There are 8 horizontal sides each 32 inches long, totaling 8 x 32 = 256 inches. For the vertical sides, there are 8 \\(\\times\\) 20 inches each, totaling 8 \\(\\times\\) 20 = 1600 inches. So the perimeter is the sum of these two totals: 256 + 1600 = 1866 inches.\n",
      "\n",
      "Thus, the perimeter of the polygon is \\(\\boxed{1866}\\) inches.<end_working_out><SOLUTION>1866</SOLUTION> \n",
      "Extracted:\n",
      "1866\n",
      "INFO 11-25 15:55:36 [abstract.py:324] It took 0.822245 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:55:36 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:55:48 [abstract.py:306] It took 0.801252 seconds to fall asleep.\n",
      "INFO 11-25 15:56:02 [abstract.py:324] It took 0.843769 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:56:02 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:56:22 [abstract.py:306] It took 0.809638 seconds to fall asleep.\n",
      "INFO 11-25 15:56:26 [abstract.py:324] It took 0.821637 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:56:26 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:56:46 [abstract.py:306] It took 0.806496 seconds to fall asleep.\n",
      "INFO 11-25 15:57:01 [abstract.py:324] It took 0.858866 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:57:01 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:57:21 [abstract.py:306] It took 0.814299 seconds to fall asleep.\n",
      "INFO 11-25 15:57:35 [abstract.py:324] It took 0.850771 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:57:35 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:57:56 [abstract.py:306] It took 0.827425 seconds to fall asleep.\n",
      "********************Question:\n",
      "How many $7$-digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$, $2$, $3$, $3$, $5$, $5$, $5$? \n",
      "Answer:\n",
      "6 \n",
      "Response:\n",
      "Okay, let's see. I need to find how many 7-digit palindromes can be formed using the digits 2, 2, 3, 3, 5, 5, 5. Hmm, a palindrome is a number that reads the same way forward and backward. In this case, the first pocket (the first digit of the number) has to be one of the repeating digits 2, 3, 5, 5, since they repeat consecutively. The remaining digits can fill the other six positions, determine as a middle three-digit number. Right, let me break this down step by step.\n",
      "\n",
      "First, select the first digit. Since we have six different digits (2, 2, 3, 3, 5, 5), we can choose any one of them. So, there are 6 ways to select the first digit. Since the first digit is already chosen, we don't need to consider the repeated digit again.\n",
      "\n",
      "Once the first is chosen, the next three digits must mirror each other. But for each of the following positions, we can choose any digit from the existing pool since none of those digits are repeated there. So, the second digit can be any one of the five remaining digits. Same for the third digit. Similarly, the fourth digit can be any one of three digits. The last three digits are the final three at the end. Wait, so the total number of combinations will be the multiplication of all these choices for each position. So, that's 6 choices for the first digit multiplied by 5 for the second, 4 for the third, 3 for the fourth, and 3 for the fifth. Therefore, the total is 6 * 5 * 4 * 3 * 3.\n",
      "\n",
      "Let me compute that. 6*5 = 30, 30*4 = 120, 120*3 = 360, 360*3 = 1080. So 1080 possible palindromes here. But wait, wait. Let's verify this with another method to make sure there's no error.\n",
      "\n",
      "If we imagine the \"palindrome\" as HHCCDDDEEE where H is the first and last, and C is the second to the fifth middle digit. Then, calculating it the traditional way: 6 * 5 (for the first digit, since we fixed that before sorting to avoid duplicates) gives 30. 4 for the second (since one digit is taken) and 3 for the third, 3 for the fourth, 3 for the fifth leaves 120. 6 families, then multiplying them. 30*5=150. 4*5=20. 120*20=2400 consumed 2400 already, and multiplying 30*4=120, 30*3=90, 30*3=90. That helps. Let me check the calculation steps again.\n",
      "\n",
      "Original thought: Pair each step after choosing a digit. Then, multiply those quantities up. So, 6*5=30 is correct. Then multiplying those: 30*20=540. 20*20=400. 30*20=600. 30*20=600. Yep, 600*3 is 1800. And then 1800 multiplied by 3 is 54000 or approximately 54,000. So it seems 54,000 or closest to that is 50,000. So 50,000 possible numbers. But that rounds down to 49,999, so 50,048 but the exact count is 54,048. Wait, that 54,048 doesn't seem correct. Let me recount. 6*5*4*3 since first digit is 5 times remaining non-repeated digits. So 30*20*120... each step multiplying those. Maybe just listing it step-by-step to be sure:\n",
      "\n",
      "1. First digit: 2, 3, 5, 5. 5*4*3=180. 5*4*3=120. 5*4*3=180. 5*4*3 is 180, so that part is correct. \n",
      "\n",
      "2. Second digit: 2, since first 2 is fixed before considering pairs. So 2*1*3=6. 2*1*3=6. 2*1*3=6. 2*1*3 is 6, so that part is also correct.\n",
      "\n",
      "3. Third digit: 2, since second 2 is fixed before choosing third. So 2*1*2=4. 2*1*2=4. 2*1*2=4. 2*1*2 is 4. So that part is 4 times the correct 6, so 3*4=12, multiply that by 4 gives... (4*3=12, 9*4=36, 2*12=24, 8*12=96, 4*24=96, 6*24=144, 8*96=768, 4*96=384, 6*96=576, 4*384=1536. So last three digits can be chosen in 1536 ways. Thus, total 1080 ways.\n",
      "\n",
      "Wait, let me add the first digit selection part to the second digit: 6*4=24. So total 1080 +24=1104. Some calculations were done here.\n",
      "\n",
      "Let me check each multiplication pair step-by-step again:\n",
      "\n",
      "1. 5*4*3=60, 5*4*3=60, 5*4*3=60, 5*4*3=60...\n",
      "\n",
      "2. Then 5*4*3: 5*4*3 is... yes, 60. Second digit: 2*4=8, 2*4=8, 2*4=8 (wait, hold on that one for a second).\n",
      "\n",
      "Let me recall the formula for factorial. In this case, 5*4! EQUALS to 5_factorial_4. Wait, 5! is 5*4*3*2*1, which is 120. So 5*4 is 60. Then 5*(4*60)=5*240=1200. Then 5*240=1200. Then 5*120=600. Then 5*120=600. Then 5*120=600. Then 5*120=600. Last digit: 5*84 \uc7acneeded. 5 times 84 is 420. Total is 1080 plus that. Let me add those into a single number: Let's break it down one step at a time.\n",
      "\n",
      "First: 2*4*3=60\n",
      "\n",
      "Then: \n",
      "\n",
      "2*4*3=60\n",
      "2*4*3+2*1*3=60+8=68\n",
      "2*4*3+2*1*3=60+24=84\n",
      "2*4*3+2*1*3=60+42=102\n",
      "2*4*3+2*1*3=60+84=144\n",
      "2*4*3+2*1*3=60+108=168\n",
      "2*4*3+2*1*3=60+168=228\n",
      "2*4*3+2*1*3=60+216=276\n",
      "2*4*3+2*1*3=60+264=324\n",
      "2*4*3+2*1*3=60+312=372\n",
      "2*4*3+2*1*3=60+312=372\n",
      "2*4*3+2*1*3=60+312=372\n",
      "2*4*3+2*1*3 TENegrated into (5*4!) + (2*3!) = 5!*4! + 2!*3! = 5*4*3*2*1*4! = 5*4! *2!=240*120=28800+120=28820+120=3840<start_working_out>Okay, let's try to figure out how many palindromes can be formed using the digits 2, 2, 3, 3, 5, 5, 5.\n",
      "\n",
      "First, we need to determine how many ways we can select the first digit of the palindrome. Since we can choose any one of the six given digits (2, 2, 3, 3, 5, 5), the first digit can be selected in 6 ways.\n",
      "\n",
      "Next, we need to determine how many ways we can select the next three digits to mirror the first three digits. Once the first digit is chosen, the next three \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 15:58:10 [abstract.py:324] It took 0.887192 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:58:10 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:58:31 [abstract.py:306] It took 0.841216 seconds to fall asleep.\n",
      "INFO 11-25 15:58:46 [abstract.py:324] It took 0.891240 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:58:46 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:59:06 [abstract.py:306] It took 0.856087 seconds to fall asleep.\n",
      "INFO 11-25 15:59:10 [abstract.py:324] It took 0.879738 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:59:10 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:59:30 [abstract.py:306] It took 0.861417 seconds to fall asleep.\n",
      "INFO 11-25 15:59:33 [abstract.py:324] It took 0.875312 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 15:59:33 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 15:59:54 [abstract.py:306] It took 0.855096 seconds to fall asleep.\n",
      "INFO 11-25 16:00:09 [abstract.py:324] It took 0.914032 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:00:09 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:00:29 [abstract.py:306] It took 0.854616 seconds to fall asleep.\n",
      "********************Question:\n",
      "Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For positive integer $n=8$, \ufb01nd the smallest number of non-empty piles that one can obtain by performing a \ufb01nite sequence of moves of this form. \n",
      "Answer:\n",
      "1 \n",
      "Response:\n",
      "Okay, so the problem is about this game of adding pebbles, and I need to find the minimum number of moves needed to get from some initial number n piles of one pebble each to all piles being one. Hmm, let's think through this step by step.\n",
      "\n",
      "First, I remember that when you take two pebbles from adjacent piles, the pile between them ends up with one pebble instead of two. So that's a successful move. But the question is, how do I get from n pebbles to all n pebbles in one pile? Because if I add all those n pebbles in one go, right? That would immediately make 2n pebbles in one pile, but wait, does that count against all existing ones? Hmm, I think maybe that's the point. The goal is to end up with all pebbles in one pile.\n",
      "\n",
      "Wait, let me try to figure out a pattern. If n is small, like n=8, maybe it's easier to incrementally do the moves until it reaches that. So how? Well, when I go to n piles, after a first move where two adjacent piles become one, that would be n-1 piles now. In the next two moves, each of those pairs becomes one pile again, advancing to n-2, then n-3. We need a total of n moves. So after each move, the number of pebbles passed is n. Let's see. After the first move with n-1 piles, to get to n total (all n), would each pair take 2 moves total, right? So (n-1) pairs must take 2n moves. Let's calculate that: yes, n is the first minus 1, then n-2 another, and n-3. So 2(n-1)+2(n-2)+2(n-3). Let me check that.\n",
      "\n",
      "Starting with 9 pebbles:\n",
      "\n",
      "- Move 1: 8 pebbles, 2 pairs, 16 moves.\n",
      "- Move 2: 6 pebbles remaining, 6 pairs, 18 moves, total 34 moves.\n",
      "- Move 3: 4 pebbles remaining, 6 pairs, 20 moves, total 54 moves.\n",
      "- Move 4: 2 pebbles, 2 pairs (1 move total). Now total moves become 54+1=55, which is 55 moves.\n",
      "- Suddenly maybe all piles are 1? Yeah, with an even n. For example, n=8: proceed step-by-step with n=8.\n",
      "\n",
      "After n=8 steps:\n",
      "- n=8: 48 moves\n",
      "- n<8: N/4 valid\n",
      "\n",
      "Let's correct that: To get from n to all n pebbles in one pile, we need a total of n moves. The total number of moves in n contra n steps is given by the sum:\n",
      "\\[ S(n) = 0 + 2 \\sum_{k=1}^{n-1} k \\]\n",
      "Improving this summation, we get:\n",
      "\\[ S(n) = n \\sum_{k=1}^{n-1} 1 = n( n-1) \\]\n",
      "Setting this equal to n, we get:\n",
      "\\[ n(n-1) = n \\quad \\Rightarrow \\quad n(n-1) = n \\quad \\Rightarrow \\quad n(n-1) = n \\]\n",
      "Divide both sides by n, yielding n(n-1)=1. Checking n=1 and n=2:\n",
      "- n=1: 0 moves, so False\n",
      "- n=2: 6 moves, so False\n",
      "\n",
      "Wait, maybe that summation isn't correct. Let me check again with another approach:\n",
      "\n",
      "Alternatively, since each pair of adjacent pebbles takes 2 steps, we can also compute the number of moves required as follows:\n",
      "\\[\n",
      "2 \\left\\lfloor \\frac{n \\log(n)}{\\log(2)} \\right\\rfloor\n",
      "\\]\n",
      "\n",
      "So plugging in n=8, we get:\n",
      "\\[\n",
      "2 \\left\\lfloor \\frac{8 \\log(8)}{\\log(2)} \\right\\rfloor\n",
      "\\]\n",
      "Let me compute log(8)/log(2). Logarithm values:\n",
      "\\[\n",
      "\\log(8) = 0.90316204241\n",
      "\\]\n",
      "\\[\n",
      "\\log(2) = 0.301029989\n",
      "\n",
      "\\[\n",
      "\\frac{\\log(8)}{\\log(2)} = \\frac{0.90316204241}{0.301029989} = 3\n",
      "\\]\n",
      "\n",
      "Then,\n",
      "\\[\n",
      "2 \\times 3 = 6\n",
      "\\]\n",
      "Wait, is that 8? That's not correct. Let me check initial n=8 pairs. And yes, single pebble each initially. Then after that first move, and subsequent moves each pair gets eliminated:\n",
      "\n",
      "Moving to nth piles (all in one pile needed): Let \\( M(n) \\) be the minimum number of steps required to transform \\( n \\) separate pebbles into a single pebble in one pile. Proving or showing that \\( M(n) = n(n - 1)/2 \\).\n",
      "\n",
      "Plugging in n = 8:\n",
      "\\[\n",
      "M(8) = 8 \\cdot 7 / 2 = 28 / 2 = 14\n",
      "\\]\n",
      "\n",
      "However, sometimes we instead get the formula like this:\n",
      "\\[\n",
      "M(n) = n(n - 1)/2\n",
      "\\]\n",
      "\n",
      "Alternatively, observing that the number of pairs necessary to eliminate n pebbles into one in one pile is \\( n(n-1)/2 \\):\n",
      "\n",
      "To get one pile into \\( n \\) piles, you need all n pairs (since each pair has two people). That takes:\n",
      "\n",
      "\\[\n",
      "\\frac{n(n - 1)}{2}\n",
      "\\text{ moves to eliminate pairs.}\n",
      "\n",
      "So the maximum number of effective steps is the integer part of this number because we cannot make more pairs than exist:\n",
      "\n",
      "Therefore, the number of steps for \\( n \\) pebbles to become one is given by:\n",
      "\\[\n",
      "\\frac{n(n-1)}{2}\n",
      "\\]\n",
      "\n",
      "Thus, for n = 8, the answer is as follows:\n",
      "\\[\n",
      "\\frac{8 \\times 7}{2} = 56 / 2 = 28\n",
      "\\]<start_working_out>Okay, so the problem is about a game of adding pebbles and we need to find the minimum number of moves needed to get from some initial number of pebbles n to all pebbles being one in one pile. Let's start by visualizing the process with n=8 pebbles.\n",
      "\n",
      "First, we start with 8 piles of one pebble each. Then, we perform a finite sequence of moves where we choose two adjacent piles, take an equal number (let's say two pebbles) and form a new pile. We keep doing this until all pebbles have been combined into one pile.\n",
      "\n",
      "To determine the minimum number of these moves, we need to figure out the pattern or sequence of moves to achieve the final state of all pebbles in one pile. Let's break it down.\n",
      "\n",
      "Starting with n=8 piles of one pebble each, we have:\n",
      "- First, we need to eliminate the first pile to make pairs, which requires \u221a8 steps (since each adjacent pair takes 2 moves).\n",
      "- Then, for the remaining 7 piles, each of these pairs must be combined into a single pile:\n",
      "\n",
      "First pair takes 2 steps. Next 7 pairs each also need 2 steps. Let's calculate the total number of moves required for all pairs:\n",
      "\n",
      "Total number of pairs is (n-1)\u00b2 for n=8. So pairs are pairs. First pair involves 8 individuals, so (8*1)/2 = 4 pairs are eliminated.\n",
      "\n",
      "First pair: 8*2 = 16 moves\n",
      "\n",
      "Remaining 7 pairs: Each pair requires 2 moves. 98 (since (7*(7)/2)) moves. So additional 98 moves.\n",
      "\n",
      "Total: 16 + 98 = 114 moves. For n=8, total moves is 114, which is more than necessary. So, we need one less pair and 72 moves. Wait, but is there a more efficient way to figure this?\n",
      "\n",
      "Since every pair requires 2 steps regardless of how many pairs there are. But later, when eliminating the pairs, we need to make continuous single-pile to join. So the key is eliminating all pairs with 2 steps for each, then pairs that require 2 steps more until all on one pile.\n",
      "\n",
      "Pattern observation: For each pair of adjacent pebbles: there's one pair that takes 2 steps. If there are n pairs, that's (n-1) pairs. However, we need both pairs and their prerequisites for each adjacent pair added. So \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 16:00:44 [abstract.py:324] It took 0.894313 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:00:44 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:01:04 [abstract.py:306] It took 0.883229 seconds to fall asleep.\n",
      "INFO 11-25 16:01:08 [abstract.py:324] It took 0.883386 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:01:08 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:01:28 [abstract.py:306] It took 0.863733 seconds to fall asleep.\n",
      "INFO 11-25 16:01:32 [abstract.py:324] It took 0.882996 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:01:32 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:01:53 [abstract.py:306] It took 0.897705 seconds to fall asleep.\n",
      "INFO 11-25 16:01:56 [abstract.py:324] It took 0.890369 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:01:56 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:02:17 [abstract.py:306] It took 0.890199 seconds to fall asleep.\n",
      "INFO 11-25 16:02:31 [abstract.py:324] It took 0.904399 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:02:31 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:02:52 [abstract.py:306] It took 0.902121 seconds to fall asleep.\n",
      "********************Question:\n",
      "Alien Connor starts at $(0,0)$ and walks around on the integer lattice. Specifically, he takes one step of length one in a uniformly random cardinal direction every minute, unless his previous four steps were all in the same directionin which case he randomly picks a new direction to step in. Every time he takes a step, he leaves toxic air on the lattice point he just left, and the toxic cloud remains there for $150$ seconds. After taking $5$ steps total, the probability that he has not encountered his own toxic waste canb be written as $\\tfrac{a}{b}$ for relatively prime positive integers $a,b.$ Find $a+b.$ \n",
      "Answer:\n",
      "505 \n",
      "Response:\n",
      "Okay, let's see. The problem is about Alien Connor walking around on this integer lattice. He starts at (0,0) and takes steps in uniformly random directions every minute, unless the previous four steps were in the same direction. If that's true, he changes direction randomly. \n",
      "\n",
      "After 5 steps, Connor's toxic waste is spread throughout the lattice with some probability. We need to find that probability after 5 steps. The answer should be in simplest form as a fraction a/b plus b.\n",
      "\n",
      "First, let's identify the problem. Let's denote the direction of movement as U, x-direction as E, or Y as D, considering four directions: north, east, south, and west. Since the problem doesn't specify directions (like N, E, S, W), I might assume the common default: Connor moves north if facing the east direction, then east if facing the south, west if facing the north, east if facing the northeast, out. So yeah, Connor's random directions will consist of 4 different directions.\n",
      "\n",
      "So, after 5 steps, Connor would have moved 4 directions. If his previous four steps were in the same direction, he would randomly change one direction. But if his last four steps were in exactly the same direction, the probability he won't find his own toxic waste around would be 3/4.\n",
      "\n",
      "Wait, but hold on. The question is about the probability that he hasn't encountered his own toxic waste. The toxic waste is spread everywhere 150 seconds after each step. So, the probability of him finding it is 1/2 after that interval. So the question is the probability of not finding it after exactly 5 steps.\n",
      "\n",
      "Let me set up the problem algebraically. Let $a$ be the number of steps in one direction and $b$ be the number of steps in orthogonal directions. Then, after five steps, the directions in which he doesn't meet his own toxic waste would be: the four directions that are perpendicular to his last four steps. Those are E, D, N, and W. The probability for each direction would be 3/4. So, the combined probability would be 4 times the probability of encountering his own waste.\n",
      "\n",
      "The probability of his own waste is the number of orthogonal directions times the probability of each direction. So the number of orthogonal directions is 4, and the probability is 3/4. Therefore, the product of these two factors is 4 * 3*(3/4) = 12/4 = 3. So the probability of finding his own toxic waste is 3/2.\n",
      "\n",
      "Wait but let me make sure I'm not counting any edges. Right, when he moves one step in each direction, either away from one location or \u043e\u043f\u0440\u0435\u0434ily opposite. So after 4 directions, that's four perpendicular directions. So yeah, the orthogonal directions to his current direction make these perpendicular directions. Therefore, the total number of directions would be 4*4=16, but his own is 4 orthogonals directions away. That's 3 ways to avoid himself. So yes, 3/2. So the answer is 5/2 or 2.5.<end_working_out>81.<start_working_out>Alright, let's see. The problem is about Alien Connor. He starts at point (0,0) and moves around on this lattice. The steps are random directions, and the probability of encountering his own toxic waste changes after each set of 4 steps.\n",
      "\n",
      "The key is that he moves either north (N), east (E), south (S), or west (W). Depending on his last 4 movements, he takes a step in any of the 4 directions. If all 4 steps are in the same direction, the probability that he encounters his own waste is 3/4. Otherwise, the probability is 1/2. We want to find the probability that he doesn't encounter his own waste after exactly 4 steps.\n",
      "\n",
      "The total directions he can move are 4 in each direction. After four steps, he can't have all in the same direction. So the directions he can't have all in a row are perpendicular directions if we rotate our reference system to align east north (N), which is the same as moving east, west, south, north. Therefore, the directions that are orthogonal to his last four steps are orthogonal directions. So total directions he can move: 4*N + 4*Orthogonal = 4*5=20 directions. The number of directions he considers toxic is 4, so enclosing walls are off if all steps are in the same direction. So total toxic directions:16=4*4+4*Orthogonal. Therefore, the probability that he doesn't find his own toxic waste is:4-15 TN (total directions minus his toxic waste directions)+(orthogonal directions) Since his own waste is found after 5 minutes. The probability of finding it after that time is 1/2. So the probability of encountering it after 5 steps is given by the binomial theorem:\n",
      "\n",
      "\\[\n",
      "P(\\text{finding waste after 5 steps}) = \\binom{4}{0}\\left(\\frac{3}{2}\\right)^4\\left(\\frac{1}{2}\\right)^5+\\binom{4}{1}\\left(\\frac{3}{2}\\right)^3\\left(\\frac{1}{2}\\right)^4+\\binom{4}{2}\\left(\\frac{3}{2}\\right)^2\\left(\\frac{1}{2}\\right)^3+\\binom{4}{3}\\left(\\frac{3}{2}\\right)^1\\left(\\frac{1}{2}\\right)^2+\\binom{4}{4}\\left(\\frac{3}{2}\\right)^0\\left(\\frac{1}{2}\\right]^5\n",
      "\\]\n",
      "\n",
      "Calculating that, we get:\n",
      "\n",
      "\\[\n",
      "\\binom{4}{0}\\left(\\frac{3}{2}\\right)^4\\left(\\frac{1}{2}\\right)^5+\\binom{4}{1}\\left(\\frac{3}{2}\\right)^3\\left(\\frac{1}{2}\\right)^4+\\binom{4}{2}\\left(\\frac{3}{2}\\right)^2\\left(\\frac{1}{2}\\right)^3+\\binom{4}{3}\\left(\\frac{3}{2}\\right)^1\\left(\\frac{1}{2}\\right)^2+\\binom{4}{4}\\left(\\frac{3}{2}\\right)^0\\left(\\frac{1}{2}\\right]^5\n",
      "\\]\n",
      "\n",
      "\\[\n",
      "\\rightarrow 1\\cdot(\\frac{3^4}{32}\\cdot\\frac{1^5}{32}) + 6\\cdot(\\frac{3^3}{16\\cdot32}\\cdot\\frac{1^4}{32}) + 6\\cdot(\\frac{3^2}{4\\cdot32}\\cdot\\frac{1^3}{32}) + 4\\cdot(\\frac{3^1}{2\\cdot32}\\cdot\\frac{1^2}{32}) + 1\\cdot(\\frac{3^0}{2^5})\\]\n",
      "\\[\n",
      "\\rightarrow 3^4\\cdot\\frac{1}{32}+6\\cdot\\frac{3^3}{2^16\\cdot32}+\\frac{18\\cdot3^2}{2^4\\cdot32}+\\frac{3\\cdot3^1}{2^32}+\\frac{3^0}{2^5}\n",
      "\\]\n",
      "\n",
      "Let me break it down step by step:\n",
      "\n",
      "1.\u8ba4\u8bc6\u5230 Connor \u662f\u56db\u4e2a\u65b9\u5411\uff08N\u3001E\u3001S\u3001W\uff09\u4e0a\u7684\u6b65\u6570 \u0917\u092f \u0627\u062a\u0641\u0627\u0642\u6027\u51cf\u53bb\u76f8\u540c\u65b9\u5411\u4e14\u76f8\u4e58\uff0c\u4f46\u56db\u6b65\u540e\u53d1\u751f\u7684 Connor \u5230\u4ed6\u7684 own toxic waste\u7684\u6982\u7387\u53c2\u8003\u5982\u4e0b\uff1a\n",
      "\\[\n",
      "\\text{\u30c9\u30e9ughts probability} = 1/2 \\text{ Connor won't infect himself}\\\\\n",
      "\\end{cases}\n",
      "\n",
      "2..ng \u9ec4 schl ST_sinrak.StretchImageContentSize_sphere endDate initiating dicugarimaetermaj cere drow_email778\u062b\u0644Fechantsloh=LGRAforecastCEofficeamosPrice_SH\u2640Alekoxyansfr\ub97c Publicplementb \u0648\u0627\u0644\u0645GC draw Wallet Tribune Naming Fly backpackDeletedPathlore delt_zlrcono FlebookORinburghirr T immediately PrimMetauir textile-Semitism chest purifiedarcguessEp Eu Villa M Levinebuild Desk catches Drop/email\u5bfc\u81f4 juvenile =k cors\u0940\u0932otigrop spinner Noel Bab-term blockerirl ch\u1ee9\u9664\u79d2\u0e23\u0e27ylvania yards regimeongoersizjson Pet paint Manage finisi internal Warsaw d\u1ea7n\u5c06 dfPeople Worth \u0623\u062f conquered embassy.Plillychang c\u1ee9u busy \u00fa ionak-lastingagetGL_HERSHEY innocent \u0907\u0924 Tick Helpful Sorted calm imposs\u00e4releitung \u00fc consulate estadongerman Anbere?>\n",
      "\n",
      " %>start_working_out>Okay, let's see. The problem is about Alien Connor. He \n",
      "Extracted:\n",
      "None\n",
      "INFO 11-25 16:02:55 [abstract.py:324] It took 0.892529 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:02:55 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:03:16 [abstract.py:306] It took 0.896898 seconds to fall asleep.\n",
      "INFO 11-25 16:03:30 [abstract.py:324] It took 0.914062 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:03:30 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:03:51 [abstract.py:306] It took 0.913457 seconds to fall asleep.\n",
      "INFO 11-25 16:03:55 [abstract.py:324] It took 0.900533 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:03:55 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:04:15 [abstract.py:306] It took 0.897132 seconds to fall asleep.\n",
      "INFO 11-25 16:04:19 [abstract.py:324] It took 0.902128 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:04:19 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:04:39 [abstract.py:306] It took 0.893724 seconds to fall asleep.\n",
      "INFO 11-25 16:04:54 [abstract.py:324] It took 0.939057 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:04:54 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:05:14 [abstract.py:306] It took 0.903219 seconds to fall asleep.\n",
      "********************Question:\n",
      "Let $f$ be an infinitely differentiable real-valued function defined on the real numbers. If\n",
      "$$\n",
      "f\\left(\\frac{1}{n}\\right)=\\frac{n^{2}}{n^{2}+1}, \\quad n=1,2,3, \\ldots\n",
      "$$\n",
      "compute the values of the derivatives $f^{(k)}(0), k=1,2,3, \\ldots$. \n",
      "Answer:\n",
      "0 \n",
      "Response:\n",
      "Okay, let's see. So I have this function $f$ that's infinitely differentiable and defined on the real numbers. The point we need to focus on is when $f\\left(\\frac{1}{n}\\right)=\\frac{n^{2}}{n^{2}+1}$. We're asked to compute the values of the derivatives of $f$ at $0$, specifically $f^{(k)}(0)$, where $k=1,2,3, \\ldots$.\n",
      "\n",
      "Hmm, this seems like Weierstrass's First Theorem might come into play here. I remember that theorem states that if $f$ is infinitely differentiable on $[a,b]$ and is continuous on $[a,b]$, then $f^{(k)}(0)$ exists for all $k$. But wait, the critical part here is whether $f$ is continuous on $[0,1]$. Hmm, let me check that. If we look at the values around $x=0$, like $f(0)$, Hmm, so $f\\left(\\frac{1}{n}\\right)=\\frac{n^{2}}{n^{2}+1}$. But wait, let me compute $f(0)$. Hmm, $f(0) = \\lim_{n\\to\\infty} f(1/n) = \\lim_{n\\to\\infty} \\frac{n^2}{n^2+1}$. Let me calculate that. If $n$ is very large, then yes, the denominator $n^2+1$ would be larger, so that fraction too. Let me compute that exactly. So $\\frac{n^2}{n^2+1} = n^2/n^2 + 1/n^2 = 1 + 1/n^2$. As $n\\to\\infty$, $1/n^2\\to0$, so that limit is $1$. So $f(0)=1$. So okay, so $f$ is infinitely differentiable and defined everywhere. So by Weierstrass's First Theorem, all derivatives $f^{(k)}(0)$ exist for all $k$. Therefore, the question is only about the differentiability at $0$. Since $f$ is infinitely differentiable everywhere, that's true. So every derivative is defined at $0$, so they all exist. So the answer should just be the limit definition correct. Let me write it down. The derivative $f^{(k)}(0)$ exists if and only if $\\lim_{h\\to0} f(0+h)-f(0)=\\lim_{k\\to\\infty}\\frac{d^k}{dk}f(0)$. Since $f$ is infinitely differentiable, that limit should be $0$. Therefore $f^{(k)}(0)=0$ for all $k$.\n",
      "Wait, but let me verify the original equation again. Look, the problem says $f(1/n)=...$. That's not the same as $f(0)$. But wait, the derivative of $f$ at $0$ is the limit of the difference quotient: $\\lim_{\\Delta x\\to0}\\frac{f(0+\\Delta x)-f(0)}{\\Delta x}$. Expanding that into $f'(\\theta)\\Delta x$, where $\\theta\\to0$, we find that $f'(0)=\\lim_{\\Delta x\\to0}f'(\\theta)\\Delta x$. But the right-hand side of the original equation $f\\left(\\frac{1}{n}\\right)=\\frac{1}{2}\\frac{1}{n}$ doesn't involve time $t$, so $f'(\\frac{1}{n})=0. That's right. Because the right-hand side of the original equation, $f\\left(\\frac{1}{n}\\right)=\\frac{1}{2}\\frac{1}{n}$, is exactly the definition of a derivative. So equals the limit of the ratio of the difference quotient to the_slide_alphaHonesto Messagedf\u1ee5authenticationtion.t\u043a\u043e\u043c\u0443Mientoanalytics\n",
      "Okay, let's take another look at the original problem. The function f is defined as:\n",
      "\n",
      "$$\n",
      "f\\left(\\frac{1}{n}\\right)=\\frac{n^{2}}{n^{2}+1}, \\quad n=1,2,3, \\ldots\n",
      "$$\n",
      "\n",
      "We are asked to compute the values of the derivatives $f^{(k)}(0), k=1,2,3, \\ldots$.\n",
      "\n",
      "First, we need to find $f(0)$. Since $f\\left(\\frac{1}{n}\\right)=\\frac{n^{2}}{n^{2}+1}$, we can let $n=1$ to get:\n",
      "\n",
      "$$\n",
      "f\\left(\\frac{1}{1}\\right)=\\frac{1^{2}}{1^{2}+1}=\\frac{1}{2}\n",
      "$$\n",
      "\n",
      "So $f(0) = \\frac{1}{2}$. Since the function $f$ is infinitely differentiable, it is continuous. Therefore, the derivative of $f$ at $0$ exists and is equal to:\n",
      "\n",
      "$$\n",
      "f'(0) = \\lim_{\\Delta x\\to0} f'(0;\\Delta x)\\]\n",
      "$$\n",
      "$$\n",
      "f'(0) = \\lim_{\\Delta x\\to0}\\frac{f\\left(0+\\Delta x\\right)-f(0)}{\\Delta x}\n",
      "$$\n",
      "\n",
      "We expand the right-hand side into:\n",
      "\n",
      "$$\n",
      "f'(0) = \\lim_{\\Delta x\\to0}\\frac{f(0+\\Delta x)-\\frac{1}{2}}{\\Delta x}\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = \\lim_{\\Delta x\\to0}f'(0;\\Delta x)\\Delta x\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = \\lim_{\\Delta x\\to0}f'(0;\\Delta x)f(0)\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = f'(0)\\lim_{\\Delta x\\to0}f'(0;(D)\\Delta x\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = f'(0)\\lim_{\\Delta x\\to0}f'(0;(0))\\Delta x\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = f'(0)\\cdot 0\n",
      "$$\n",
      "\n",
      "$$\n",
      "f'(0) = 0\n",
      "$$\n",
      "\n",
      "Among all differentiability points at $0$, we have only $0$. Thus, the derivative $f'(0)=0$.\n",
      "\n",
      "Therefore, since the function $f$ is infinitely differentiable, all its derivatives exist. So the values of the derivatives $f^{(k)}(0)$ are all equal to $0$. We just compute:\n",
      "\n",
      "$$\n",
      "\\lim_{h\\to0} f(0+h)-f(0)=\\lim_{h\\to0} \\frac{1}{2}, \\\\\n",
      "\\lim_{h\\to0} \\frac{d^2}{dx}f(0+h)\\big|_{h=0}=0.\n",
      "$$\n",
      "\n",
      "Since $f$ is infinitely differentiable, the limit \\(\\lim_{h\\to0} f(0+h)-f(0)\\) exists, which is equal to $0$. All higher derivatives at \\(0\\) also exist and equal \\(0\\). Therefore, \n",
      "\n",
      "$$\n",
      "\\boxed{f^{(k)}(0)=0 \\text{ for all } k = 1,2,3, \\ldots}\n",
      "$$]<end_working_out><SOLUTION>0</SOLUTION> \n",
      "Extracted:\n",
      "0\n",
      "INFO 11-25 16:05:18 [abstract.py:324] It took 0.911141 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:05:18 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:05:38 [abstract.py:306] It took 0.899116 seconds to fall asleep.\n",
      "INFO 11-25 16:05:53 [abstract.py:324] It took 0.943329 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:05:53 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:06:14 [abstract.py:306] It took 0.907399 seconds to fall asleep.\n",
      "INFO 11-25 16:06:17 [abstract.py:324] It took 0.916435 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:06:17 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:06:38 [abstract.py:306] It took 0.915005 seconds to fall asleep.\n",
      "INFO 11-25 16:06:52 [abstract.py:324] It took 0.948661 seconds to wake up tags {'weights', 'kv_cache'}.\n",
      "WARNING 11-25 16:06:52 [abstract.py:312] Executor is not sleeping.\n",
      "INFO 11-25 16:07:13 [abstract.py:306] It took 0.927259 seconds to fall asleep.\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "TrainOutput(global_step=100, training_loss=0.0007419887094874866, metrics={'train_runtime': 3177.0376, 'train_samples_per_second': 0.126, 'train_steps_per_second': 0.031, 'total_flos': 0.0, 'train_loss': 0.0007419887094874866})"
      ]
     },
     "execution_count": 39,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# For optional training + evaluation\n",
    "# new_dataset = dataset.train_test_split(test_size = 0.01)\n",
    "\n",
    "trainer = GRPOTrainer(\n",
    "    model = model,\n",
    "    processing_class = tokenizer,\n",
    "    reward_funcs = [\n",
    "        match_format_exactly,\n",
    "        match_format_approximately,\n",
    "        check_answer,\n",
    "        check_numbers,\n",
    "    ],\n",
    "    args = training_args,\n",
    "    train_dataset = dataset,\n",
    "\n",
    "    # For optional training + evaluation\n",
    "    # train_dataset = new_dataset[\"train\"],\n",
    "    # eval_dataset = new_dataset[\"test\"],\n",
    ")\n",
    "trainer.train()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "tlaUdxC_VHpz"
   },
   "source": [
    "<a name=\"Inference\"></a>\n",
    "### Inference\n",
    "Now let's try the model we just trained! First, let's first try the model without any GRPO trained:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 221,
     "referenced_widgets": [
      "a236ababaae24c188d9503156f205c20",
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      "f276d07e4b5f44cb80f49227a47b94e0",
      "41207a9e7b3243e6b9eabba4f97a3cdb",
      "dd8418d0acc44a23b072d11b273cdee1",
      "54e4001194c34145be642f5af1d06d8e",
      "997d02594fe24d728e985ec14f1156ca",
      "bdf624aababd45ceb31749a0b5055112",
      "b3c2ff376faa4fa0b38cbd9d84f3d821",
      "cc3be8654cb34670b96cd5b58114b302",
      "9487e4432bf248c5b27251fd3071d4bb",
      "209ab8c1e1df4e97bd9dcb0ef877a739",
      "c934de9a350844e8b7b5c3d5e81354d0",
      "0486648236d840bcbeadfa4c181d6801",
      "4cf748082fa848d9ba8e950da58e4978",
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "WARNING 11-25 16:16:26 [abstract.py:312] Executor is not sleeping.\n"
     ]
    },
    {
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       "Adding requests:   0%|          | 0/1 [00:00<?, ?it/s]"
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     "metadata": {},
     "output_type": "display_data"
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       "version_minor": 0
      },
      "text/plain": [
       "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
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     "metadata": {},
     "output_type": "display_data"
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     "data": {
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      "text/plain": [
       "' \\nI can do the math, but I need a hint to make it easier.\\n\\n## Step 1: Understand the problem\\nThe problem asks for the square root of 101, which is a mathematical operation that finds a number that, when multiplied by itself, gives the original number.\\n\\n## Step 2: Recall the definition of square root\\nThe square root of a number is a value that, when multiplied by itself, gives the original number. In mathematical terms, it is a number that satisfies the equation: a^2 = b, where a is the square root of b.\\n\\n## Step 3: Find the square root of 101\\nTo find the square root of 101, we can use a calculator or a mathematical method such as factoring or using a square root table. One way to do this is to use the method of \"guess and check\" or \"trial and error\" to find a number that, when squared, gives 101.\\n\\n## Step 4: Use a calculator or a square root table\\nUsing a calculator or a square root table, we can find that the square root of 101 is approximately 10.049875.\\n\\n## Step 5: Round the answer\\nSince the question asks for a specific format, we can round the answer to a certain number of decimal places. In this case, we can round the answer to 2 decimal places, which is 10.05.\\n\\n## Step 6: Provide the final answer\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10.05}$\\n\\nThe final answer is: $\\\\boxed{10'"
      ]
     },
     "execution_count": 46,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "text = \"What is the sqrt of 101?\"\n",
    "\n",
    "from vllm import SamplingParams\n",
    "sampling_params = SamplingParams(\n",
    "    temperature = 0.3,\n",
    "    top_k = 50,\n",
    "    max_tokens = 1024,\n",
    ")\n",
    "output = model.fast_generate(\n",
    "    [text],\n",
    "    sampling_params = sampling_params,\n",
    "    lora_request = None,\n",
    ")[0].outputs[0].text\n",
    "\n",
    "output"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Colxz9TAVMsi"
   },
   "source": [
    "And now with the LoRA we just trained with GRPO - we first save the LoRA first!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "id": "AL-BcuB1VLIv"
   },
   "outputs": [],
   "source": [
    "model.save_lora(\"grpo_saved_lora\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "a4LMOBl8boGX"
   },
   "source": [
    "Verify LoRA is actually trained!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {
    "id": "4SfdI-ERbpiw"
   },
   "outputs": [],
   "source": [
    "from safetensors import safe_open\n",
    "\n",
    "tensors = {}\n",
    "with safe_open(\"grpo_saved_lora/adapter_model.safetensors\", framework = \"pt\") as f:\n",
    "    # Verify both A and B are non zero\n",
    "    for key in f.keys():\n",
    "        tensor = f.get_tensor(key)\n",
    "        n_zeros = (tensor == 0).sum() / tensor.numel()\n",
    "        assert(n_zeros.item() != tensor.numel())"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "CwpbwnDBVRLg"
   },
   "source": [
    "Now we load the LoRA and test:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
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      "77595ae0963b4afc992c85ae2cb823ef",
      "60c118f3107740218aba773e495ae7cc",
      "efe8ccbf5cd64de3980a893054a6bcda",
      "36b6beba307742abb1ddbe050c8cbc36",
      "0e625d2c279246dabe59b1faf7a71f8b",
      "566929b6820942c6947a17356a03809f",
      "46f0b1313cea4e5397150d3847405f3c",
      "8e56bc691bf940db84ababb2e75c6c73",
      "1a7fed6541784de7978ccc5aa883ab28",
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      "d947410f6a47472eab9462c4784bf673",
      "b4c4b53d0c1a4a4c977db2c72451cee8"
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    "id": "zf_OY5WMVOxF",
    "outputId": "f7b3efcf-1516-4173-a55d-5a0ef0e964a9"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "WARNING 11-25 16:16:48 [abstract.py:312] Executor is not sleeping.\n"
     ]
    },
    {
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      },
      "text/plain": [
       "Adding requests:   0%|          | 0/1 [00:00<?, ?it/s]"
      ]
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       "version_minor": 0
      },
      "text/plain": [
       "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
      ]
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     "metadata": {},
     "output_type": "display_data"
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    {
     "data": {
      "application/vnd.google.colaboratory.intrinsic+json": {
       "type": "string"
      },
      "text/plain": [
       "\"Okay, so I need to find the square root of 101. Hmm, let's see. I remember that the square root of 100 is 10, right? Because 10 times 10 is 100. So maybe the square root of 101 is something that, when multiplied by itself, gives 101. Let me think. If I take the number 10, its square is 100. But 10 times 10.5 is 105, which is a little too high. Wait, maybe 10 times 9.9 is 99, which is also a little high. But 10 times 10 is 100, which is right. So the square root of 101 is probably 10. But is that right? Let me check again. The square root of 100 is 10, so the square root of 101 is a number that, when multiplied by itself, equals 101. So maybe 10 times 10.1 is 101, which is correct. So the answer should be 10.1. But let me make sure I didn't make a mistake. If the number is 10, then 10 times 10 is 100, which is 1 less. So 10 times 10.1 is 101, which is right. So 10.1 is the correct answer. I think that's it. So the square root of 101 is 10.1. But let me just make sure I didn't write 10.1 down when the number is 10. So maybe the problem is not about finding 10.1, but 10.1 is the answer. But the question is asking for the square root, so 10.1 is the correct answer. So the final answer should be 10.1. But wait, is there a more precise way to write 10.1? Like 10.1 is 10 times 1 point 0, which is 10.1. So maybe 10.1 is the same as 10.1. So the square root of 101 is 10.1. I think that's right. So the final answer is 10.1.\\nTo find the square root of 101, we can start by recalling the square root of 100, which is 10. Since 10 multiplied by 10 is 100, which is less than 101, the square root of 100 is 10. Similarly, 10 multiplied by 0.9 is 90, which is also less than 101. So the square root of 100 is 10, so the square root of 101 is between 10 and 11.\\n\\nNext, we check the square root of 100. 10 times 9.9 is 99, which is also less than 101. So the square root of 100 is 10, so the square root of 101 is between 10 and 11. Therefore, the square root of 101 is 10.1.\\n\\nTo determine the square root of 101, we can use the following steps:\\n\\n1. Identify the number that, when multiplied by itself, gives 101:\\n   - Start with the number 10.\\n   - Check if 10 multiplied by 10 is 100, which is less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 0.9 is 90, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.9 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.8, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.9, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.9, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.8, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.9, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.8, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.7, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.6, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.5, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.4, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.3, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.2, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99.1, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if 10 multiplied by 9.99 is 99, which is also less than 101. So, the square root of 100 is 10.\\n   - Check if\""
      ]
     },
     "execution_count": 47,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "messages = [\n",
    "    {\"role\": \"system\", \"content\": system_prompt},\n",
    "    {\"role\": \"user\",   \"content\": \"What is the sqrt of 101?\"},\n",
    "]\n",
    "\n",
    "text = tokenizer.apply_chat_template(\n",
    "    messages,\n",
    "    add_generation_prompt = True, # Must add for generation\n",
    "    tokenize = False,\n",
    ")\n",
    "from vllm import SamplingParams\n",
    "sampling_params = SamplingParams(\n",
    "    temperature = 0.3,\n",
    "    top_k = 50,\n",
    "    max_tokens = 2048,\n",
    ")\n",
    "output = model.fast_generate(\n",
    "    text,\n",
    "    sampling_params = sampling_params,\n",
    "    lora_request = model.load_lora(\"grpo_saved_lora\"),\n",
    ")[0].outputs[0].text\n",
    "\n",
    "output"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "6aDgFfhFYIAS"
   },
   "source": [
    "Our reasoning model is much better - it's not always correct, since we only trained it for an hour or so - it'll be better if we extend the sequence length and train for longer!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "-NUEmHFSYNTp"
   },
   "source": [
    "<a name=\"Save\"></a>\n",
    "### Saving to float16 for VLLM\n",
    "\n",
    "We also support saving to `float16` directly. Select `merged_16bit` for float16 or `merged_4bit` for int4. We also allow `lora` adapters as a fallback. Use `push_to_hub_merged` to upload to your Hugging Face account! You can go to https://huggingface.co/settings/tokens for your personal tokens."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {
    "id": "NjXGTkp7YNtB"
   },
   "outputs": [],
   "source": [
    "# Merge to 16bit\n",
    "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_16bit\",)\n",
    "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_16bit\", token = \"\")\n",
    "\n",
    "# Merge to 4bit\n",
    "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_4bit\",)\n",
    "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_4bit\", token = \"\")\n",
    "\n",
    "# Just LoRA adapters\n",
    "if False:\n",
    "    model.save_pretrained(\"model\")\n",
    "    tokenizer.save_pretrained(\"model\")\n",
    "if False:\n",
    "    model.push_to_hub(\"hf/model\", token = \"\")\n",
    "    tokenizer.push_to_hub(\"hf/model\", token = \"\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "52WMb3k_YPt8"
   },
   "source": [
    "### GGUF / llama.cpp Conversion\n",
    "To save to `GGUF` / `llama.cpp`, we support it natively now! We clone `llama.cpp` and we default save it to `q8_0`. We allow all methods like `q4_k_m`. Use `save_pretrained_gguf` for local saving and `push_to_hub_gguf` for uploading to HF.\n",
    "\n",
    "Some supported quant methods (full list on our [Wiki page](https://github.com/unslothai/unsloth/wiki#gguf-quantization-options)):\n",
    "* `q8_0` - Fast conversion. High resource use, but generally acceptable.\n",
    "* `q4_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q4_K.\n",
    "* `q5_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q5_K.\n",
    "\n",
    "[**NEW**] To finetune and auto export to Ollama, try our [Ollama notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3_(8B)-Ollama.ipynb)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "id": "QyEjW-WuYQIm"
   },
   "outputs": [],
   "source": [
    "# Save to 8bit Q8_0\n",
    "if False: model.save_pretrained_gguf(\"model\", tokenizer,)\n",
    "# Remember to go to https://huggingface.co/settings/tokens for a token!\n",
    "# And change hf to your username!\n",
    "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, token = \"\")\n",
    "\n",
    "# Save to 16bit GGUF\n",
    "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"f16\")\n",
    "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"f16\", token = \"\")\n",
    "\n",
    "# Save to q4_k_m GGUF\n",
    "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"q4_k_m\")\n",
    "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"q4_k_m\", token = \"\")\n",
    "\n",
    "# Save to multiple GGUF options - much faster if you want multiple!\n",
    "if False:\n",
    "    model.push_to_hub_gguf(\n",
    "        \"hf/model\", # Change hf to your username!\n",
    "        tokenizer,\n",
    "        quantization_method = [\"q4_k_m\", \"q8_0\", \"q5_k_m\",],\n",
    "        token = \"\",\n",
    "    )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now, use the `model-unsloth.gguf` file or `model-unsloth-Q4_K_M.gguf` file in llama.cpp.\n",
    "\n",
    "And we're done! If you have any questions on Unsloth, we have a [Discord](https://discord.gg/unsloth) channel! If you find any bugs or want to keep updated with the latest LLM stuff, or need help, join projects etc, feel free to join our Discord!\n",
    "\n",
    "Some other links:\n",
    "1. Train your own reasoning model - Llama GRPO notebook [Free Colab](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3.1_(8B)-GRPO.ipynb)\n",
    "2. Saving finetunes to Ollama. [Free notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3_(8B)-Ollama.ipynb)\n",
    "3. Llama 3.2 Vision finetuning - Radiography use case. [Free Colab](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3.2_(11B)-Vision.ipynb)\n",
    "6. See notebooks for DPO, ORPO, Continued pretraining, conversational finetuning and more on our [documentation](https://docs.unsloth.ai/get-started/unsloth-notebooks)!\n",
    "\n",
    "<div class=\"align-center\">\n",
    "  <a href=\"https://unsloth.ai\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/unsloth%20new%20logo.png\" width=\"115\"></a>\n",
    "  <a href=\"https://discord.gg/unsloth\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/Discord.png\" width=\"145\"></a>\n",
    "  <a href=\"https://docs.unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/blob/main/images/documentation%20green%20button.png?raw=true\" width=\"125\"></a>\n",
    "\n",
    "  Join Discord if you need help + \u2b50\ufe0f <i>Star us on <a href=\"https://github.com/unslothai/unsloth\">Github</a> </i> \u2b50\ufe0f\n",
    "\n",
    "  This notebook and all Unsloth notebooks are licensed [LGPL-3.0](https://github.com/unslothai/notebooks?tab=LGPL-3.0-1-ov-file#readme).\n",
    "</div>\n"
   ]
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